\documentstyle[amssymbo,tr_def,12pt]{article} \sloppy \renewcommand{\baselinestretch}{1.25} \topmargin 1.3cm \oddsidemargin 1cm \parindent 0.5cm \parskip 0.25cm \textwidth 13.6cm \textheight 17.8cm \newenvironment{mini1}{\begin{minipage}[t]{5.5cm}}{\end{minipage}} \newcommand{\be}{\begin{equation}} \newcommand{\ee}{\end{equation}} \newcommand{\bea}{\begin{eqnarray}} \newcommand{\eea}{\end{eqnarray}} \newcommand{\mb}{\mbox{\ }} \newcommand{\vs}{\vspace} \newcommand{\bc}{\begin{center}} \newcommand{\ec}{\end{center}} \newcommand{\ben}{\begin{enumerate}} \newcommand{\een}{\end{enumerate}} \newcommand{\bfi}{\begin{figure}} \newcommand{\efi}{\end{figure}} \newcommand{\ra}{\rightarrow} \newcommand{\al}{\alpha} \newcommand{\sg}{\Sigma} \newcommand{\bt}{\beta} \newcommand{\gm}{\gamma} \newcommand{\dt}{\delta} \newcommand{\ep}{\epsilon} \newcommand{\vr}{\varphi} \newcommand{\pr}{\partial} \newcommand{\lam}{\lambda} \newcommand{\iy}{\infty} \newcommand{\si}{\sigma} \newcommand{\la}{\leftarrow} \newcommand{\ovl}{\overline} \newcommand{\unl}{\underline} \newcommand{\R}{\Bbb{R}} \newcommand{\N}{\Bbb{N}} \newcommand{\bq}{\begin{quote}} \newcommand{\eq}{\end{quote}} \newcommand{\bqu}{\begin{quotation}} \newcommand{\equ}{\end{quotation}} \newenvironment{emphit}{\begin{itemize}}{\end{itemize}} \newcommand{\bemp}{\begin{emphit}} \newcommand{\eemp}{\end{emphit}} %a,b,c'li denklemler icin % \newcounter{sayi} %\renewcommand{\theequation}{\Alph{sayi}.\arabic{equation}} %\newcommand{\bseq}{\begin{subequations} \begin{eqalignno}} %\newcommand{\eseq}{\end{eqalignno} \end{subequations}} %\begin{document} %\setcounter{sayi}{1} %1.1., 2.1.'li denklem %\renewcommand{\theequation}{\arabic{section}.\arabic{equation}} %\newcommand{\bseq}{\begin{subequations} \begin{eqalignno}} %\newcommand{\eseq}{\end{eqalignno} \end{subequations}} %\begin{list}{}{\usecounter{enumi}\setlength{\leftmargin}{1.5cm} %\setlength{\labelwidth}{1.2cm}} %\item[ ] %\end{list} %\pagestyle{empty} \begin{document} % ekte bas¨ var. \bc {\large \bf On The Commutant of The Ideal Centre} \ec \bc {\em ˛afak Alpay, Bahri Turan} \ec {\bf Abstract:}{\small The centre $Z(E)$ of a Riesz space $E$ is the collection of all operators $T$ on $E$ that satisfy $-\lam I \leq T \leq \lam I$ for some $\lam \in \R$, where $I$ is the identity operator. We study the commutant $Z(E)_C$ of $Z(E)$ in the order bounded operators on $E$. If $E$ is a Riesz space with topologically full centre, we identify $Z(E)_C$ with the orthomorphisms of $E$.} \mb \vs{0.1cm}\\ \bc {\bf Preliminaries} \ec If $E$ is a Riesz space $E^\sim$, the order dual of $E$ will be the Riesz space of all order bounded linear functionals on $E$. Riesz spaces considered in this note are assumed to have separating order duals. $Z(E)$ will denote the ideal centre, Orth $(E)$, will denote the orthomorphisms of $E$. If $E$ is a topological Riesz space $E'$ will denote continuous dual of $E$. When $T: E \ra F$ is an order bounded operator between two Riesz spaces, the adjoint of $T$ carries $F^\sim$ into $E^\sim$ and it will be denoted by $T^\sim$. In all undefined terminology concerning Riesz spaces we will adhere to the definitions in [1],[6] and [11]. When the order dual $E^\sim$ separates the points of the Riesz space $E$, an order bounded operator $T: E \ra E$ is an orthomorphism if and only if its adjoint $T^\sim : E^\sim \ra E^\sim$ is an orthomorphism. Moreover, the operator $\psi: T \ra T^\sim$ from Orth $(E)$ into Orth $(E^\sim)$ is a Riesz homomorphism [1]. The image under $\psi$ of the centre $Z(E)$ will be denoted by $Z^\sim (E)$. $Z^\sim (E)$ is a Riesz subspace of $Z(E^\sim)$. \newpage {\bf Definition:} A Riesz space $E$, with separating order dual $E^\sim$, is said to have topologically full centre if, for each pair $x,y$ in $E$ with $0\leq y \leq x$, there exists a net $(\pi_\al)$ in $Z(E)$ with $0 \leq \pi_\al \leq I$ for each $\al$, such that $\pi_\al x\ra y$ in $\si (E, E^\sim)$. Banach lattices with topologically full centre were initiated in [10]. These spaces were also studied in [2], [3] and [8]. The class of Riesz spaces and the class of Banach lattices that have topologically full centre are quite large. $\si$-Dedekind complete Riesz spaces have topologically full centres. However, not all Riesz spaces have topologically full centres. {\bf Example:} [12] Let $E$ be the Riesz space of piecewise affine, continuous functions on $[0,1]$. Considered as a sublattice of $C[0,1]$, $E$ is cofinal in $C[0,1]$. That is to say for each $0\leq f $ in $C[0,1]$ there exists $e \in E$ with $f\leq e$. Hence each positive functional on $E$ can be extended to a positive functional on $C[0,1]$ by Theorem 83.15 in [11]. Therefore, we have $E^\sim = C[0,1]^\sim$. % Hence we have %$E^\sim = C[0,1]^\sim$. But $Z(E)$ consists only of multiples of identity and $Z(E)$ is not topologically full. To see this let $T \in Z(E)$ and $f \in E$. Let $x \in [0,1]$ be arbitrary and define $f_x =f-f(x)\cdot 1$. $f_x \in E$ and $|T f_x| \leq \lam |f_x|$ for some $\lam \in \R_+$. $|Tf_x| (x)=0$ as $f_x (x)=0$. Hence $(Tf)(x) =(T1)(x)f(x)$ for all $x \in [0,1]$. $T(1)$ is a constant function on $[0,1]$. If $T(1)$ were not a constant function then $(T1)(x)= ax+b$ on some of open interval. Then we would have $(T(T1))(x)=a^2x^2+2abx + b^2$ on some interval. This would violate the linearity of $T(T1)$ on that interval. Hence $(Tf)(x)=(T1)(x) f(x)$ so that $Tf=\al_Tf$ with $\al_T =T(1)$. There exists an AM-space $E$ whose centre consists of multiples of the identity therefore $Z(E)$ is not topologically full [9]. As Banach lattices are relatively uniformly complete and have separating order duals, we see that relative uniform completess does not, in general, imply topological fullness of the centre. We now give an example of a Riesz space which has topologically full centre but is not Dedekind or $\sigma$-Dedekind complete. {\bf Example:} Let $X$ be a topological space. We denote by $C_b (X)$ the AM-space of all bounded, continuous real valued functions on $X$. It is well-known that if $X$ is also normal $C_b(X)$ is $\sigma$-Dedekind (Dedekind) complete if and only if each open $F_\sigma$-subset of $X$ has open closure (each subset of $X$ has open closure). However, unlike these completeness properties, $C_b(X)$ always have topologically full centre for a normal topological space $X$. Let us note that $C_b(X)$ can be identified with its centre. Clearly, each $f \in C_b (X)$ defines an operator $T(g)= f\cdot g$, $g \in C_b (X)$, in the ideal centre. For a proof that this correspondence is an onto lattice isomorphism we refer the reader to Theorem 8.27 in [1]. To see that $C_b (X)$ has topologically full centre, we let $0\leq y \leq x$ in $C_b(X)$ and $0<\epsilon$ be arbitrary. Let $D= \{t \in X : \epsilon \leq x(t)\}$. The function $z(t)= y(t)/x(t)$ is continuous and bounded on the closed set $D$. Tietze extension theorem allows us to extend $z=z(t)$ to an element in $C_b(X)$. We denote the extension by $z=z(t)$ again and consider the operator $T$ defined by $z$. Clearly, $\mid \mid y-T(x)\mid \mid < \ep $ and $C_b(X)$ has topologically full centre. %{\bf Example:} It is well known that if $E$ %is an $f$-algebra, $(E^\sim)^\sim_n$, %equipped with the Areas multiplication is %again an $f$-algebra [5]. This is achieved in three %steps: given $a,b \in E, f \in E^\sim$ and %$F,G \in (E^\sim)^\sim_n$ %one defines $f\cdot a \ep E^\sim$, $F\cdot f \ep E^\sim$ and %$F\cdot G \ep (E^\sim)^\sim_n$ by the equations %$(f\cdot a) (b)= f(ab)$, $(F\cdot f)(a)=F(f\cdot a)$ %and $(F\cdot G)(f)=F(G\cdot f)$. $(E^\sim)^\sim$ and %$(E^\sim)^\sim_n$ coincide if $E$ has a unit %element [5]. As special case we have $C(X)'' = %(C(X)')'_n$ for all topological spaces $X$. Let %$X$ be a compact Hausdorff space. Let $E$ be $C(X)$ %with usual norm and order. $E''$ is a unital %$f$-algebra with multiplication as defined above. %We can identify $E''$ with Orth $(E'')= Z(E'')$ %through the map $F\ra \pi_F$ where %$\pi_F (G) =F\cdot G$ for each $F,G \in E''$. Let $x,y$ %be elements of $E$ such that $0\leq y \leq x$. %Let $\hat{x}, \hat{y}$ be the images of $x$ and $y$ in %$E''$. %Then $0\leq \hat{y} \leq \hat{x}$ in $E''$. Since %$E''$ is Dedekind complete there exists an %orthomorphism $T$ on $E''$ with $T(\hat{x}) =\hat{y}$ [1]. %By considering the orthomorphism $|T| \wedge I$ %if necessary, we can suppose $0\leq T \leq I$. %Thus these exists $F \in E''$ with $\pi_F =T$. %In particular, $\pi_F (\hat{x})= \hat{x}\cdot F %= F\cdot \hat{x} =T(\hat{x}) =\hat{y}$, %$E$ is $|\si| (E'', E')$ %dense in $E''$ %by Theorem 11.16 in [1]. %Thus there exists a net $(e_\al)$ in $E$ %such that $e_\al \ra F$ in $|\si| (E'', E')$. %Since lattice operations are continuous for %$|\si | (E'', E')$ and $0 \leq F \leq 1$, by %considering $|e_\al| \wedge 1$ if necessary, we %can suppose $0\leq e_\al \leq 1$ for all $\al$. %Let $f \in E'$. Then $f^+ (e_\al \cdot x) =f^+ (x\cdot e_\al)= %f^+ \cdot x (e_\al) \stackrel{\al}{\ra} f^+ \cdot x(F)$ as %$f^+ \cdot x \in E'$. On the other hand, since $f^+ \cdot x(F)= %f^+ (\hat{x} \cdot F) =\hat{x} \cdot F(f^+)$, we have %$(\hat{e}_\al \cdot \hat{x}) = \widehat{e_\al \cdot x} \stackrel{\al}{\ra} %\hat{x} \cdot F$ in $\si (E'', E')$. That is to say %$\hat{e}_\al \cdot \hat{x} (f) =\hat{e}_\al (\hat{x}\cdot f) %\stackrel{\al}{\ra} \hat{x} \cdot %F(f)=f (y)$. %Thus $\hat{x} \cdot f(e_\al) \stackrel{\al}{\ra} f(y)$ or %$\hat{x} (f\cdot e_\al) \stackrel{\al}{\ra} f(y)$. Since %$\hat{x} (f\cdot e_\al) =f\cdot e_\al (x)= f(e_\al \cdot x)$ for %each $\al$, we obtain $f(e_\al \cdot x) \ra f(y)$ %for each $f \in E'$. $e_\al \cdot x$ is the image %of $x$ under the orthomorphism $\pi_{e_\al}$ for %each $\al$. Since $f \in E'$ is arbitrary, we %conclude that $E$ has a topologically full centre. {\small We wish to thank the referee cordially for his valuable advice and suggestions.} \bc {\bf Main Results} \ec Order bounded maps on the Riesz space $E$ will be denoted by $L_b (E)$. $Z(E)_C$ will denote the commutant of $Z(E)$ in $L_b(E)$. That is, $Z(E)_C =\{ T \in L_b (E): T\pi = \pi T$ for each $\pi \in Z(E)\}$. Since $E^\sim$ is assumed to separate the points of $E,E$ is Archimedean. It follows that the Riesz space Orth $(E)$ under composition is an Archimedean $f$-algebra and therefore it is commutative. Hence Orth$(E) \subset Z(E)_C$. In this note we study the relationship between Orth$(E)$ and $Z(E)_C$. The notion of relative uniform convergence for lattices is due to Kantorovitch [5]. As is done by Gordon in [4], the notion of relative uniform convergence can be generalized so as to apply to spaces that are not vector lattices. {\bf Definiton:} A sequence $(x_n)$ in an ordered vector space $E$ is said to converge relative uniform to $x \in E$ if there exists $e \in E_+$ and $(\lam_n) \subset \R_+$ with $\lam_n \ra 0$ such that, for all $n \in \N$, \[-\lam_n e \leq x-x_n \leq \lam_n e.\] A subset $Y$ of $E$ is called relatively uniformly closed if whenever a sequence $(x_n)$ in $Y$ converges relative uniform to $x, x$ must also be in $Y$. $Z(E)_C$ is a relatively uniformly closed subset of $L_b(E)$. Let $(T_n)$ be a sequence in $Z(E)_C$ which converges relative uniform to $T$. %Then it is easy to see that %$T \in L_b (E)$. To show $T \in Z(E)_C$, it is enough to show $T \pi = \pi T$ whenever $0\leq \pi \leq I$. Let us choose $(\lam_n) \subset \R_+$ with $\lam_n \ra 0$ and $U \in L(E)_+$ with $-\lam_n U \leq T-T_n \leq \lam_n U$ for each $n$. Then for each $\pi \in Z(E)_+$ with $0\leq \pi \leq I$, we have $-2\lam_n U \leq T\pi -\pi T =T\pi -\pi T + \pi T_n - T_n \pi \leq 2\lam_n U$ for each $n$. Since $L_b(E)$ is Archimedean, we have $T\pi = \pi T$ and $Z(E)_C$ is relatively uniformly closed. Let $T_\al$ and $T$ be order bounded operators on the Riesz space $E$. If $\sup_\al \{ T_\al x\}= Tx$ for each $x \in E_+$, we write $T=\sup_\al T_\al$. Since $Z(E)$ is positively generated and consists of order continuous operators on $E$, it is easy to see that if $T=\sup_\al T_\al$ with $T_\al \in Z(E)_C$ for each $\al$ then $T\in Z(E)_C$. Let $E, F$ and $G$ be Riesz spaces. A bilinear map $\phi: E \times F \ra G$ is called a bi-lattice homomorphism if for each $f \in F_+,$ the maps $y \ra \phi (y,f)$ of $E$ into $G$ and the maps $F\ra G$ defined by $g\ra \phi (x,g)$ for each $x \in E_+$ are lattice homomorphisms. In particular, we shall consider the bilinear map $\phi: E \times E^\sim \ra Z(E)'$ defined by $\phi (x,f)=\mu_{x,f}$, where $\mu_{x,f} (\pi) =f(\pi (x))$ for $\pi \in Z(E)$. For each $x \in E_+$, the maps $f \ra \phi (x,f)$ and for each $f \in E_+^\sim$ the maps $x\ra \phi (x,f)$ are positive and we have $| \phi (x,f)| \leq \phi (|x|,|f|)$ for each $(x,f)$ in $E \times E^\sim$. If $E$ has topologically full centre, we can say more about the positivity of the map $\phi$. {\bf Lemma:} Let $E$ be a Riesz space with separating order dual $E^\sim$. If $E$ has topologically full centre, then $\phi: E \times E^\sim \ra Z(E)'$ is a bi-lattice homomorphism. {\bf Proof:} First, we show $\phi (x, \cdot): E^\sim \ra Z(E)'$ is a lattice homomorphism for each $x \in E_+$. Positivity of the map $\phi (x, \cdot): E^\sim \ra Z(E)'$ implies that $\phi (x,f)^+ \leq \phi (x, f^+)$ for each $x \in E_+$ and $f \in E^\sim$. Let $\pi \in Z(E)_+,$ and $x \in E_+$ be arbitrary, then \[ \phi (x, f^+)(\pi) =\mu_{x, f^+} (\pi) =f^+ (\pi x) =\sup \{ f(y): 0\leq y \leq \pi (x)\}.\] For each $y$ with $0\leq y \leq \pi (x)$, we can find $(\pi_\al)$ in $Z(E)$ such that $0\leq \pi_\al \leq I$ for each $\al$ and $\pi_\al (\pi x) \ra y$ in $\si (E, E^\sim)$ as $E$ has topologically full centre. As $0\leq \pi_\al (\pi x) \leq \pi (x)$ for each $\al$, we have \[f(\pi_\al (\pi x)) \leq \phi (x,f)(\pi_\al \pi)\leq \phi (x,f)^+ (\pi)\quad \mbox{for each}\quad \al.\] Hence $f(y) \leq \phi (x,f)^+ (\pi)$ for each $y$ with $0 \leq y \leq \pi (x)$. Taking supremum over all such $y$, we obtain $\phi (x, f^+) \leq \phi (x, f)^+$ for an arbitrary $f \in E^\sim$. Now we show that $\phi (\cdot, f): E \ra Z(E)'$ is a lattice homomorphism for each $f \in E^\sim_+$. It is enough to show that $\phi (x, f) \wedge \phi (y,f) =0$ for each $x,y$ in $E$ satisfying $x \wedge y =0$. As $I$, the identity operator on $E$, is a strong order unit in $Z(E)$ it suffices to show \[ \phi (x, f) \wedge \phi (y,f) (I) = (\mu_{x,f} \wedge \mu_{y,f} ) (I)=0.\] It is well-known that \begin{eqnarray*} (\mu_{x,f} \wedge \mu_{y,f}) (I)& =& \inf \{ \mu_{x,f} (\pi_1) + \mu_{y,f} (\pi_2) : 0\leq \pi_1, \pi_2 \in Z(E), \quad \pi_1 + \pi_2 =I \}\\ & = & \inf \{ f(\pi_1 x) + f(\pi_2 x): 0\leq \pi_1, \pi_2 \in Z(E), \quad \pi_1 + \pi_2 = I\}. \end{eqnarray*} Let $z= x+y$. Let $I_x, I_y$ and $I_z$ be, respectively, the order ideals generated by $x,y$ and $z$. Then $I_z$ is actually the order direct sum of $I_x$ and $I_y$ by Theorem 17.6 in [10] . We denote by $P$ the order projection of $I_z$ onto $I_x$. Let $J$ be the restriction to $I_z$ of order bounded functionals on $E$. Then $J$ is an order ideal in $I_z^\sim$ because if $f \in I_z^\sim$ satisfies $0\leq f \leq g_{|_{I_z}}$ for some $g \in E^\sim$, then $f$ has an extention to a positive functional on $E$ by Theorem 2.3 in [1]. The adjoint $P^\sim : I_z^\sim \ra I_z^\sim$ of $P$ satisfies $0\leq P^\sim \leq I$ and as a consequence we obtain $P^\sim (J) \subset J$. As a result of these simple observations we obtain that the pair $< I_z, J>$ constitutes a Riesz pair and $P: I_z (\si (I_z, J))\ra I_z (\si (I_z, J))$ continuous. Since $0 \leq P(z)\leq z,$ there exists $(\pi_\al)$ in $Z(E)$, $0\leq \pi_\al \leq I$ such that $\pi_\al(z) \ra P(z)$ in $\si (E, E^\sim)$ and therefore in $\si (I_z, J)$. We have $P(\pi_\al z)= \pi_\al (Pz)= \pi_\al (x)$, and the continuity of $P$ now yields $\pi_\al x \ra x$ in $\si (I_z, J)$. Since we have $\pi_\al (z) =\pi_\al (x) + \pi_\al (y)$ for each $\al$, we have $\pi_\al (y)\ra 0$ in $\si (I_z, J)$. As $\mu_{x,f} \wedge \mu_{y,f} (I) \leq f(( I-\pi_\al) (x) + \pi_\al (y))$ for each $\al$, we obtain $(\mu_{x,f} \wedge \mu_{y,f})(I) \leq \lim_\al f((I-\pi_\al)(x) + \pi_\al (y) )=0$ which completes the proof. \hspace*{4.0cm} $\hfill \Box$ A well-known Theorem of H.Nakano (cf. [1], Theorem 8.3) states that if $E$ has the principal projection property then the band preserving operators are precisely the ones that commute with order projections. Thus, for Riesz spaces having the principal projection property, $Z(E)_C$ is precisely Orth $E$. Since $Z(E)_C =$ Orth $E$ for a Dedekind complete Riesz space $E$, we have $Z(E)_C$ is relatively uniformly complete. Thus it is natural to ask whether $Z(E)_C$ is relatively uniformly complete whenever $E$ is. Let $\{ \pi_n\}$ be a relatively uniformly Cauchy sequence of positive operators in $Z(E)_C$. That is to say, there exists $T \in L_b (E)_+$ such that for every $\ep >0$, there exists an integer $N=N(\ep)$ with \bc $-\ep T \leq \pi_n -\pi_m \leq \ep T$ for all $n,m \geq N.$ \ec Hence, we have $-\ep T(x) \leq \pi_n (x) - \pi_m (x) \leq \ep T(x)$ for each $x \in E_+$ and all $n,m \geq N$. Since $E$ is relatively uniformly complete and $E_+$ is relatively uniformly closed, there exists a unique $\pi_0 (x) \ep E_+$ such that $\pi_n (x) \ra \pi_0 (x)$ relatively uniformly. $\pi_0$ extends uniquely to a positive operator on $E$. Using that $\pi_n$ is order bounded for each $n$, it is straightforward to prove that $\pi_0$ is order bounded. To show $\pi_0 \in Z(E)_C$ it is enough to show $\pi_0$ commutes with each $0 \leq T \ep Z(E)$, which follows easily from the fact that a positive operator is sequentially relatively uniformly continuous. When $E$ is Dedekind complete, commutativity of a positive operator with the operators in the ideal centre implies that it is an interval preserving operator. Let $0\leq S \ep Z(E)_C$ be arbitrary, then for each $z \in E^+$, $S[0,z] \subseteq [0, Sz]$. Thus, it is enough to verify $[0,Sz] \subseteq S[0,z]$. Let $0\leq y \leq Sz$ be arbitrary. Then there exists some $\pi \in Z(E)$, $0\leq \pi \leq I$ satisfying $\pi (Sz) = S(\pi z)=y$. Thus $y \in S[0,z]$. %The next result generalizes this to Riesz spaces with %topologically full centre. {\bf Proposition:} Let $E$ be a Riesz space with separating order dual $E^\sim$ and topologically full centre. Then $Z(E)_C =$ Orth $E$. {\bf Proof:} Let $T \in Z(E)_C$ be arbitrary. It is easy to see that $\mu_{Tx,f} = \mu_{x,\tilde{T}f}$ for each $x \in E$ and $f \in E^\sim$. On the other hand if $x \perp y $ in $E$, then positivity of the map $\phi: E \times E^\sim \ra Z(E)'$ implies that $| \mu_{x,f} | \leq \mu_{|x|,|f|} \leq \mu_{|x|, |f| \vee |g|}$ and $|\mu_{y,g}| \leq \mu_{|y|,|g|} \leq \mu_{|y|, |f|\vee |g|}$ for each $f,g$ in $E^\sim$. Hence $0 \leq |\mu_{x,f} |\wedge |\mu_{y,g}|\leq \mu_{|x|, |f|\vee |g|} \wedge \mu_{|y|,|f|\vee |g|}= \mu_{|x|\wedge |y|, |f|\vee |g|} =0$ by the lemma. Thus we obtain that, if $x \perp y$ in $E$, then $\mu_{x,f} \perp \mu_{y,g}$ in $Z(E)'$ for each $f,g$ in $E^\sim$. In particular, we have $\mu_{x, \tilde{T} f} \perp \mu_{y,f}$ for each $f \in E^\sim$. Hence $\mu_{Tx, f} \perp \mu_{y,f}$ for each $x,y$ in $E$ with $x\perp y$ and $f \in E^\sim$. Since $E^\sim$ separates the points of $E$, the result $Tx \perp y$ follows from the fact that $x\ra \phi (x,f)$ is a lattice homomorphism for each $f \in E^\sim_+$. \hfill $\Box$ Results show that for a Banach lattice with order continuous norm we have $Z(E) = Z(E^\sim) = Z(E')$, where $E'$ denotes the topological dual [7]. The next result characterizes Riesz spaces for which $Z(E) = Z(E^\sim)$. {\bf Proposition:} Let $E$ be a Dedekind complete Riesz space. Then $E^\sim = E^\sim_n$ if and only if the adjoint map $\psi: Z(E) \ra Z(E^\sim)$ is surjective. {\bf Proof:} Suppose $E^\sim = E^\sim_n$. Then the topology $\si (E, E^\sim)$ is order continuous. Therefore $E$ is an order ideal of $(E^\sim)^\sim_n$ by Theorem 11.13 in [1]. Let $T \in Z(E^\sim)$ be arbitrary. Then $T^\sim \in Z((E^\sim)^\sim_n)$ and it leaves $E$ invariant. Thus we conclude that $T^\sim |_E$ is in $Z(E)$. It is straightforward to see that $(T^\sim |_E )^\sim =T$. Thus $\psi$ is surjective. Suppose $\psi$ is surjective. To show that $E^\sim = E^\sim_n$ it is enough to prove that each band in $E^\sim$ is $\si (E^\sim,E)$-closed by Theorem 11.10 in [1]. Let $B$ be an arbitrary band in $E^\sim$ and $P$ be the band projection on $B^d$. Then $P \in Z(E^\sim)$. By hypothesis there exists $P_1 \in Z(E)$ with $P^\sim_1 = P$. As $P: (E^\sim, \si (E^\sim, E) ) \ra (E^\sim, \si (E^\sim , E) )$ continuous, $B= P^{-1} (0)$ is $\si (E^\sim, E)$ closed. \hfill $\Box$ Let us recall that $Z^\sim (E)$ is the image of $Z(E)$ under the adjoint map $\psi: Z(E) \ra Z(E^\sim)$. {\bf Proposition:} Let $E$ be a Riesz space with topologically full centre. Then $Z^\sim (E)_C =$ Orth $(E^\sim)$. {\bf Proof:} $E^\sim$ has topologically full centre as it is Dedekind complete. Therefore $Z(E^\sim)_C=$ Orth $(E^\sim)$ by previous proposition. $Z^\sim (E) \subset Z(E^\sim) \subset$ Orth $(E^\sim)$ implies that Orth $ (E^\sim) \subset Z^\sim (E)_C$. Thus it is enough to show that $Z^\sim (E)_C \subset$ Orth $(E^\sim)$. Let $T$ be an order bounded operator on $E^\sim$ that commutes with every operator $\tilde{\pi}, \pi \in Z(E)$. Suppose $f \perp g$ in $E^\sim$. We claim $Tf \perp g$ in $E^\sim$. Let us consider the bilinear map $\Psi : E^\sim \times E^{\sim\sim} \ra Z(E^\sim)'$ defined by $\Psi (f, F)= \mu_{f, F}$ where $\mu_{f,F} (\pi) = F(\pi f)$ where $F \in E^{\sim\sim}$, $f \in E^\sim$ and $\pi \in Z(E^\sim)$. $E^\sim$ has topologically full centre and the bilinear map $\Psi: E^\sim \times E^{\sim \sim} \ra Z(E^\sim)'$ has similar properties to those of $\phi: E \times E^\sim \ra Z(E)'$ defined earlier. In particular, it is a bi-lattice homomorphism, and if $x \ra \hat{x}$ is the canonical embedding of $E$ into $E^{\sim\sim}$, then $\mu_{x,f} (\pi) = \mu_{f, \hat{x}} (\tilde{\pi})$ for each $x \in E, f \in E^\sim$ and $\pi \in Z(E)$. Since $f \perp g$, we have $\mu_{f,F} \perp \mu_{g,G}$ for arbitrary $F,G$ in $E^{\sim\sim}$. In particular, for $F= T^\sim \hat{x}$ and $G=\hat{x}$ we obtain $\mu_{f,\tilde{T} \hat{x} } \perp \mu_{g,\hat{x}}$ in $Z(E^\sim)'$. We let $Z^\sim (E)^\circ$ be the polar of $Z^\sim (E)$ in $Z(E^\sim)'$. Then $(Z^\sim (E)^\circ)^d$ is a projection band in $Z(E^\sim)'$. $P$ will denote the band projection on $(Z^\sim (E)^\circ)^d$. Let us consider $Z(E^\sim)'|_{Z^\sim (E)}$ in $Z^\sim (E)'$. $Z(E^\sim)' |_{Z^\sim (E)}$ is the class of order bounded functionals on $Z^\sim (E)$ that have order bounded extentions to $Z(E^\sim)$ and is an order ideal of $Z(E^\sim)'$. Therefore $Z(E^\sim)' |_{Z^\sim (E)}$ is a Riesz subspace of $Z(E^\sim)'$. The map $P(\mu) \ra \mu |_{Z^\sim (E)}$ is a Riesz isomorphism of $(Z^\sim (E)^\circ)^d$ onto $Z(E^\sim)' |_{Z^\sim (E)}$. By Theorem 7.3 in [1], it is enough to show that it is a positive bijection with positive inverse. Suppose $\mu_1 |_{Z^\sim(E)} = \mu_2 |_{Z^\sim(E)}$ for some $\mu_1, \mu_2\in Z(E^\sim)'$. Then $\mu_1 - \mu_2 |_{Z^\sim (E)}=0$ so that $(\mu_1 -\mu_2) \in Z^\sim (E)^\circ$. Hence $P(\mu_1 -\mu_2)= P(\mu_1) - P(\mu_2)=0$ or $P(\mu_1) = P(\mu_2)$. Positivity of $P$ $\mu \ra \mu |_{Z^\sim (E)}$ is clear. Surjectivity of $\mu \ra \mu |_{Z^\sim (E)}$ is also clear. Positivity of the inverse follows from Theorem 2.3. in [1]. Since a positive orthomorphism is a lattice homomorphism $P(\mu_{f, \tilde{T} \hat{x} }) \perp P(\mu_{g,\hat{x}})$ follows from $\mu_{f,\tilde{T}\hat{x} } \perp \mu_{g,\hat{x}}$. Therefore, we have $\mu_{f,\tilde{T}\hat{x}} |_{Z^\sim (E)} \perp \mu_{g,\hat{x}} |_{Z^\sim (E)}$. Since $T \in Z^\sim (E)_C$. $T\tilde{\pi} =\tilde{\pi} T$ for each $\pi \in Z(E)$ and we have $\mu_{f,\tilde{T}\hat{x}} |_{Z^\sim (E)} = \mu_{Tf, \hat{x}} |_{Z^\sim (E)}$. Therefore $\mu_{Tf,\hat{x}}|_{Z^\sim (E)} \perp \mu_{g,\hat{x}} |_{Z^\sim (E)}$. As $\mu_{Tf, \hat{x}} (\tilde{\pi}) = \mu_{x, Tf} (\pi)$ for each $\pi \in Z(E)$, $f \in E^\sim$ and $x \in E$ it follows that $\mu_{x,Tf} \perp \mu_{x,g}$ in $Z(E)'$. $\phi: E \times E^\sim \ra Z(E)'$ is a bi-lattice homomorphism implies that $\mu_{x,|Tf|\wedge |g|} =0$ in $Z(E)'$ for each $x \in E_+$. Evaluating at $I \in Z(E)$ we see that $(|Tf|\wedge |g|) (x)=0$ for each $x \in E_+$. Hence $|Tf| \wedge |g|=0$ as $E^\sim$ separates the points of $E$. This shows $Tf \perp g$ in $E^\sim$ and $T \in$ Orth $(E^\sim)$. \hfill $\Box$ Commutant of $Z^\sim (E)$ seems to be related to another subalgebra of $L_b (E^\sim)$ which we now define. Let $W(E^\sim)$ be defined as \bc $W(E^\sim) =\{ T \in L_b (E^\sim): T(I) \subseteq I$ for each $\si (E^\sim, E)$-closed ideal $I$ in $E^\sim \}$. \ec When $E$ has a topologically full centre, the class of $\si (E^\sim, E)$-closed ideals of $E^\sim$ coincide with the $\si (E^\sim, E)$ closed $Z(E)$-submodules of $E^\sim$. Thus $W(E^\sim)$ can be described as the order bounded operators on $E^\sim$ that leave invariant each $\si (E^\sim,E)$-closed $Z(E)$-submodule of $E^\sim$. The next result describes elements of $W(E^\sim)$ in terms of the bilinear map $(x,f) \ra \mu_{x,f}$ of $E\times E^\sim \ra Z(E)'$. {\bf Proposition:} Let $E$ be a Riesz space with separating order dual $E^\sim$ and topologically full centre. An operator $T \in L_b (E^\sim)$ is in $W(E^\sim)$ if and only if $\mu_{x,f}=0$ implies $\mu_{x,Tf} =0$ for $x \in E, f \in E^\sim$. {\bf Proof:} Let $T \in W(E^\sim)$ and $\mu_{x,f}=0$ for some $x \in E, f \in E^\sim$. We have $(Z(E)x)^0=I_x^0$ for a Riesz space with topologically full centre. Hence, $f \in (I_x)^0$ which is a $\si (E^\sim, E)$-closed band of $E^\sim$. Therefore $Tf \in (Z(E)x)^0$. That is to say, $Tf(\pi x)=\mu_{x,Tf} (\pi)=0$ for each $\pi \in Z(E)$ or $\mu_{x,Tf}=0$ as claimed. Let now $I\subset E^\sim$ be a $\si (E^\sim, E)$-closed ideal of $E^\sim$. $(^0I)$, polar of $I$ taken in $E$, is an ideal in $E$. We wish to show $T(I) \subseteq I$. Let us choose $f \in I_+$ and $x \in (^0I)_+$ and observe that $\mu_{x,f}(\pi)=0$ for each $\pi \in Z(E)$. It follows that $\mu_{x,f} (\pi)=0$ for each $x \in (^0I)$ and $f \in I$. Then, by hypothesis, $\mu_{x,Tf}=0$, $x \in (^0I), f \in I$. Thus $Tf \in (^0I)^0$ and $Tf \in I$ by the Bipolar theorem. \mb \hspace*{4.0cm}$\hfill \Box$ $\sigma (E^\sim, E)$-closed ideals of $E^\sim$ are bands (cf. Theorem 106.1 in [11]). Therefore for a Riesz space with topologically full centre, we have $Z^\sim (E)_C \subseteq W(E^\sim)$. In general, an arbitrary band in $E^\sim$ may not be $\si (E^\sim, E)$-closed. A necessary and sufficient condition for this to happen is $E^\sim = E^\sim_n$. Thus, $W(E^\sim)=Orth (E^\sim)$ whenever $E^\sim=E^\sim_n$. \bc {\bf REFERENCES} \ec \begin{enumerate} \item[{[1]}] Aliprantis, C.D., Burkinshaw, O., {\em Positive Operators}.\\ Academic Press, London 1985. \item[{[2]}] Alpay, ˛., Turan, B., {\em On surjectivity of the Arens homomorphism}.\\ Turkish J. Mathematics 20(3), (1996)369-377. \item[{[3]}] Alpay, ˛., Turan, B., {\em On $f$-modules}.\\ Rev. Roumain de Math. Pures et App. 40(1995) 233-241. \item[{[4]}] H.Gordon, {\em Relative Uniform Convergence}\\ Math. Ann. 153 (1964), 418-427. \item[{[5]}] L.V.Kantorovitch, {\em Sur les propri$\acute{e}$tes des espaces semi-ordenn$\acute{e}$s lin$\acute{e}$ares},\\ C.R.Acad. Sci. Paris 202 (1936), 813-816. \item[{[6]}] Luxemburg, W. A.J., Zaanen, A.C., {\em Riesz spaces I}.\\ North-Holland, Amsterdam, 1971. \item[{[7]}] Orhon, M., {\em The ideal centre of the dual of a Banach lattice}.\\ Techn. Univ. Darmstadt, Math. Dept. Preprint Series No. 1182, 1988. \item[{[8]}] Orhon, M., {\em The orthocentre and the ideal centre of the order dual of a Riesz space}.\\ Middle East Techn. Univ. Math. Dept. Preprint Series 3 (1989). \item[{[9]}] Rugy, A.G., {\em La structure id$\acute{e}$ale des $M$-espaces}\\ J.Math. Pures at Appl. 51 (1972), 331-373. \item[{[10]}] Wickstead, A.W., {\em Extremal Structure of cones of Operators}.\\ Quart. J. Math. Oxford 32(2), 239-253, (1981). \item[{[11]}] Zaanen, A.C., {\em Riesz spaces II}.\\ North-Holland, Amsterdam, 1983. \item[{[12]}] Zaanen, A.C., {\em Examples of Orthomorphisms}.\\ J.Approximation Theory 13(1979), 192-204. \end{enumerate} \begin{tabular}{lllllllllllll} ˛afak ALPAY &&&&&&&&&&&& Bahri TURAN\\ Mathematics Department &&&&&&&&&&&& Mathematics Department\\ Middle East Technical University &&&&&&&&&&&& Gazi University\\ 06531 Ankara, Turkey &&&&&&&&&&&& Ankara, Turkey\\ e-mail: safak@rorqual.cc.metu.edu.tr &&&&&&&&&&&&\\ \end{tabular} \end{document}