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\begin{document}
\title[Algebraic disjointness preserving operators]{A strongly diagonal 
power of algebraic \linebreak[1]
order bounded disjointness preserving operators}

\author{Karim Boulabiar}
\address{IPEST, Universit\'{e} de Carthage, BP 51, 2070-La Marsa, Tunisia}
\email{karim.boulabiar@ipest.rnu.tn}

\author{Gerard Buskes}
\address{Department of Mathematics, University of Mississippi,  MS 38677}
\email{mmbuskes@olemiss.edu}

\author{Gleb Sirotkin}
\address{Department of Mathematics, Northern Illinois University, DeKalb, IL 
60115}
\email{sirotkin@math.niu.edu}

\thanks{The first and the second authors gratefully acknowledge support from
the NATO Collaborative Linkage Grant \#PST.CLG.979398. The second author 
also acknowledges support from the Office of Naval
Research Grant \#N00014-01-1-0322}

\subjclass[2000]{Primary 47B65, 06F20, 06F25}

\keywords{Algebraic, disjointness preserving, locally algebraic, minimal 
polynomial, orthomorphism, strongly diagonal}

\date{June 18, 2003}

\commby{Svetlana Katok}

\begin{abstract}
An order bounded disjointness preserving operator $T$ on an Archi\-medean 
vector
lattice is algebraic if and only if the restriction of $T^{n!}$ to the vector
sublattice generated by the range of $T^{m}$ is strongly diagonal, where $n$
is the degree of the minimal polynomial of $T$ and $m$ is its
`\textit{valuation}'.
\end{abstract}

\maketitle

\section{Introduction}

Consider a square matrix $T$ for which on every row there is at most one
nonzero entry. Let $n$ be the degree of its minimal polynomial and let $m$ be
its `\textit{valuation}', i.e. the multiplicity of $0$ as a root of that
minimal polynomial. Then $T^{n!}$ is diagonal, when restricted to the range of
$T^{m}$. The latter looks surprising and one suspects that the result is
known, but we have not been able to locate a reference for it. In this paper 
we
offer a wide ranging generalization of this matrix result. The condition above
simply states that the matrix represents an operator that preserves
disjointness in the pointwise ordering. The question arises naturally, as to
whether general operators on vector lattices that preserve disjointness behave
in a similar fashion. For obvious reasons, when leaving the domain of 
finite-dimensional spaces, some form or another of continuity is 
reasonably imposed
on the operators considered. Thus we study order bounded disjointness
preserving operators on Archimedean vector lattices. Such operators exhibit a
remarkable diversity. Weighted composition operators, and the classical shift
are all order bounded disjointness preserving operators, while their theory
motivates research for stochastic and doubly stochastic operators. From the
previous examples, a shift operator behaves differently from the matrix case,
whichever diagonal is considered. However, there is a concept of
diagonal such that, surprisingly, order bounded disjointness preserving
operators that satisfy a polynomial equation do behave like the matrix case.
Indeed, the algebraic orthomorphisms serve the role of diagonals. This brings
us to the main topic of our paper, algebraic order bounded disjointness
preserving operators.

Over the last two decades order bounded disjointness preserving operators have
been studied in various directions, such as, their multiplicative
representations on functions spaces (\cite{AK}, \cite{ABN}, \cite{BBH}, 
\cite{J}), their spectral theory (\cite{A}, \cite{AH}, \cite{M-N}), and 
their
different polar decompositions (\cite{AK}, \cite{BB}, \cite{GH}, 
\cite{M-N}). In
this paper we are interested in algebraic order bounded disjointness
preserving operators and their connection with what we will call
\textit{strongly diagonal operators}, i.e. step functions of the identity
operator \cite{AB}. Our main result announces that for every algebraic order
bounded disjointness preserving operator $T$ on an Archimedean vector lattice,
the restriction of $T^{n!}$ to the vector sublattice generated by the range of
$T^{m}$ is a strongly diagonal operator, where $n$ is the degree of the
minimal polynomial of $T$ and $m$ is its `\textit{valuation}'. Also, we show
that the strongly diagonal operators are precisely the algebraic
orthomorphisms. The latter explains how our study ties in with the recent
paper \cite{LPS} by Luxemburg, de Pagter and Schep, who for Dedekind complete
vector lattices, define the order projections of order bounded operators onto
the band of central orthomorphisms to be their diagonals. Finally, we point
out that the study of algebraic and locally algebraic operators was initiated
by Kaplansky \cite{K}, to which we refer for more about these operators.

As a consequence, we find that the absolute value of an algebraic order
bounded disjointness preserving operator is algebraic as well, which contrasts
with the absolute value of finite rank or compact operators.

Full proofs of the results of this note will appear elsewhere. We use the
recent book \cite{AA}, as well as \cite{AB} and \cite{M-N} as a starting point
and we refer the reader to these monographs for unexplained
terminology and notation.

\section{Strongly diagonal operators and algebraic orthomorphisms}

Throughout this section, $L$ denotes an Archimedean vector lattice. Though we
assume $L$ to be a real vector lattice, the results in this paper are valid
for complex vector lattices as well. Denote by $\mathrm{Orth}\left(  L\right)
$ the lattice ordered algebra of all orthomorphisms on $L$ and by
$\mathrm{Orth}_{a}\left(  L\right)  $ the set of algebraic orthomorphisms on
$L$. The latter has the structure of a lattice ordered algebra itself, as is
further explained in the next theorem.

\begin{theorem}
\label{21}An operator on $L$ is strongly diagonal if and only if it is an
algebraic orthomorphism.\textit{\ }In particular,\textit{\ }$\mathrm{Orth}%
_{a}\left(  L\right)  $ is a lattice ordered subalgebra of\textit{\ }%
$\mathrm{Orth}\left(  L\right)  $\textit{.}
\end{theorem}

The composition of two algebraic orthomorphisms is again an algebraic
orthomorphism. The latter is not the case in general for algebraic order
bounded disjointness preserving operators, though the composition of two order
bounded disjointness preserving operators is again an order bounded
disjointness preserving operator. A counterexample is provided by the two
algebraic order bounded disjointness preserving operators $R$ and $S$ on the
vector lattice $C\left(  \left[  0,1\right]  \right)  $ of all real-valued
continuous functions on $\left[  0,1\right]  $ defined by
\[
R\left(  f\right)  \left(  x\right)  =w\left(  x\right)  f\left(  1-x\right)
\text{\hspace{0.2cm}and\hspace{0.2cm}}S\left(  f\right)  \left(  x\right)
=f\left(  1-x\right)
\]
for every $f\in C\left(  \left[  0,1\right]  \right)  $ and $x\in\left[
0,1\right]  $, where $w\left(  x\right)  =1+x$ if $x\in\left[  0,1/2\right]  $
and $w\left(  x\right)  =9\left(  2-x\right)  ^{-1}/4$ if $x\in\left[
1/2,1\right]  $.

From the start of the study of algebraic operators by Kaplansky in \cite{K},
locally algebraic operators were a constant companion. In particular,
Kaplansky's result that the two notions coincide on Banach spaces has been
influential. In the following we will denote by $\mathrm{Orth}_{\ell a}\left(
L\right)  $ the set of all locally algebraic orthomorphisms on $L$. We give a
characterization of the latter set next.

\begin{corollary}
\label{22}An orthomorphism on $L$ is locally algebraic if and only if its
restriction to every principal band of $L$ is strongly diagonal. In
particular,\textit{\ }$\mathrm{Orth}_{\ell a}\left(  L\right)  $ is a lattice
ordered subalgebra of $\mathrm{Orth}\left(  L\right)  $\textit{.}
\end{corollary}

Whereas algebraic orthomorphisms are strongly diagonal, the same is not true
for locally algebraic orthomorphisms, even if $L$ is Dedekind complete.
Indeed, consider the Dedekind complete vector lattice $c_{00}$ of all real
sequences that are zero outside a finite set and the orthomorphism $T$ defined
on $c_{00}$ by $T\left(  u\right)  =\left(  nu_{n}\right)  _{n\in\mathbb{N}}$
for every $u=\left(  u_{n}\right)  _{n\in\mathbb{N}}\in c_{00}$. One can prove
that $T$ is locally algebraic, but not strongly diagonal. The observation that
needs to be made here is that the vector lattice $c_{00}$ is not Kaplansky
complete, a concept that is introduced by us next.

\begin{definition}
\label{23}We say that $L$ is \textit{Kaplansky complete} if, for every
infinite countable set $\mathcal{F}$ of $L$, there exists $f\in L$ and an
infinite subset $\mathcal{G}$ of $\mathcal{F}$ such that $\inf\left(
f,g\right)  \neq0$ for every $g\in\mathcal{G}$.
\end{definition}

Noticing that every Banach lattice, every $\sigma$-laterally complete vector
lattice, and every vector lattice with weak order unit is Kaplansky complete,
we next observe that Definition \ref{23} is crucial for understanding when
every locally algebraic orthomorphism is strongly diagonal.

\begin{theorem}
\label{24}The equality $\mathrm{Orth}_{\ell a}\left(  L\right)  
=\mathrm{Orth}%
_{a}\left(  L\right)  $ holds $($and hence every locally algebraic
orthomorphism on $L$ is strongly diagonal$)$ if and only if $L$\ is 
Kaplansky complete.
\end{theorem}

On the other hand, for $\sigma$-Dedekind complete vector lattices, the
condition that every orthomorphism is strongly diagonal is very strong as we
can see next.

\begin{corollary}
\label{25}Assume that $L$ is $\sigma$-Dedekind complete. Then\textit{\ every
orthomorphism on }$L$\textit{\ is strongly diagonal if and only if }%
$L$\textit{\ is finite-dimensional. In particular, every order bounded
disjointness preserving operator on }$L$\textit{\ is strongly diagonal if and
only if }$L$\textit{\ is finite-dimensional.}
\end{corollary}

Without the assumption that $L$ is $\sigma$\textit{-}Dedekind complete,
Corollary \ref{25} need not hold. Indeed, for the $\sigma$\textit{-}Dedekind
complete vector lattice $L$ of all real-valued continuous piecewise affine
functions on $\left[  0,1\right]  $ we have that $\mathrm{Orth}\left(
L\right)  =\mathbb{R}.I$, where $I$ is the identity operator of $L$. Hence,
all orthomorphisms are strongly diagonal, while $L$ is infinite-dimensional.
Also, we cannot replace `strongly diagonal'\ by `locally algebraic'. It
suffices to see that $\mathrm{Orth}\left(  c_{00}\right)  $ coincides with the
vector lattice of all real sequences and thus all of the orthomorphisms on
$c_{00}$ are locally algebraic.

\section{Algebraic order bounded disjointness preserving operators\\ and
strongly diagonal operators}

In this section, instead of orthomorphisms, we consider the more general
setting of order bounded disjointness preserving operators. In fact, this
section includes the central result of our paper. Its proof is based on the
existence of invariant principal order ideals, the Kakutani representation
theorem and the representation of order bounded disjointness preserving
operators on spaces of the type $C\left(  \Omega\right)  $ as weighted
composition operators \cite{A}. From now on, $T$ denotes an order bounded
disjointness preserving operator on the Archimedean vector lattice $L$. Let us
recall that the `\textit{valuation}' of a polynomial is the multiplicity of
$0$ as a root of that polynomial.

\begin{theorem}
\label{31}The order bounded disjointness preserving operator $T$ is
algebraic\ if and only if there exist natural numbers $m$\ and $n$, with 
$n>m$, 
such that the restriction of $T^{n!}$ to the vector sublattice of $L$
generated by the range of $T^{m}$ is strongly diagonal\textit{.} Furthermore,
if $T$\ is algebraic, then $n$\ $($respectively, $m)$\ can be chosen as the the
degree $($respectively, the `valuation'\,$)$\ of the minimal polynomial of~$T$.
\end{theorem}

It follows that the absolute value of an algebraic order bounded disjointness
preserving operator is algebraic as well. This seems far from obvious without
the representation in Theorem \ref{31} and contrasts with the fact that the
absolute value of a finite rank operator need not be a finite rank operator
\cite{AA}. We recall here that the absolute value of an order bounded
disjointness preserving operator exists and is a positive disjointness
preserving operator \cite{M}. If we make some additional assumptions, we can
improve the conclusion of Theorem \ref{31} considerably. We give two examples
of such additional conditions, each of them is of independent interest for,
e.g., spectral theory.

\begin{corollary}
\label{32}Assume that $T$ is surjective or injective. Then $T$\ is algebraic
if and only if there exists $n\in\mathbb{N}$ such that $T^{n!}$ is strongly
diagonal. Moreover, $n$\ can be chosen to be the degree of the minimal
polynomial of $T$.
\end{corollary}

\begin{corollary}
\label{33}If $T\ $is order continuous and algebraic with order dense range in
$L$, then $T^{n!}$ is strongly diagonal, where $n$ is the degree of the minimal
polynomial of $T$. If $L$ is in addition Dedekind complete, then, even without
the condition of order continuity, there exists $R\in\mathrm{Orth}(L)$ and
$S\in Z(L)$ such that $R=ST^{n!}$, where $Z\left(  L\right)  $ denotes the
centre of $L$.
\end{corollary}

We remark, at the end of this paper, that the finite-dimensional version of
Theorem \ref{31} was actually the starting point of this paper.

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\end{document}