The semilattice, box, and lattice-tensor products in quantum logic
Boris Ischi
Given two complete atomistic lattices $\l_1$ and $\l_2$, we define a set $\sep=\sep(\l_1,\l_2)$ of complete atomistic lattices by means of three axioms (natural regarding the description of {separated} quantum compound systems), or in terms of a universal property with respect to a given class of bimorphisms. We prove that $\sep$ is a complete lattice. The bottom element $\l_1\aerts\l_2$ is the {separated product} of Aerts. For atomistic lattices with $1$ (not complete), $\l_1\aerts\l_2\cong\l_1\square\l_2$ the {box product} of Gr\"atzer and Wehrung, and, in case $\l_1$ and $\l_2$ are moreover coatomistic, $\l_1\aerts\l_2\cong\l_1\boxtimes\l_2$ the {lattice tensor product}. The top element $\l_1\otf\l_2$ is the (complete) join-semilattice tensor product of Fraser, which is isomorphic to the tensor products of Chu and Shmuely. With some additional hypotheses on $\l_1$ and $\l_2$ (true if $\l_1$ and $\l_2$ are moreover orthomodular with the covering property), we prove that $\sep$ is a singleton if and only if $\l_1$ or $\l_2$ is distributive, if and only if $\l_1\otf\l_2$ has the covering property. Our main result reads: $\l\in\sep$ admits an orthocomplementation if and only if $\l=\l_1\aerts\l_2$. For $\l_1$ and $\l_2$ moreover irreducible, we characterize the automorphisms of each $\l\in\sep$ in terms of those of $\l_1$ and $\l_2$. At the end, we construct an example $\l_1\ots\l_2$ in $\sep$ which has the covering property.
SLTPinQL3.tex
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\newcommand{\abs}[1]{\lvert#1\rvert}
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\def\s{\Sigma}
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\def\Sep{\mathcal{S}}
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\def\mo{\mathsf{MO}}
\def\h{\mathcal H}
\def\pro{\mathsf{P}}
\def\prs{\s_{_\Downarrow}}
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\def\pr{\pi^{-1}}
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\def\ots{\,{\Downarrow\kern-1.08em\bigcirc}}
\def\otssub{{\Downarrow\kern-.63em\bigcirc}}
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\def\aerts{\,{\land\kern-.85em\bigcirc}}
\def\aertssub{{\rm{\land\kern-.65em \bigcirc}}}
\def\aertsbig{{\rm{\land\kern-.74em \bigcirc}}}
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\def\otf{\,{\vee\kern-.85em\bigcirc}}
\def\otfsub{{\vee\kern-.65em\bigcirc}}
\def\otfbig{{\vee\kern-.74em\bigcirc}}
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\def\otchu{{\rm{\otimes_{_{\hspace{-3.2mm} Chu}}}\hspace{-0.7mm}}}
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\def\otaerts{{\rm{\otimes_{_{\hspace{-4mm} Aerts}}}\hspace{-0.7mm}}\,}
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\begin{document}
\title{The semilattice, box, and lattice-tensor products
in quantum logic}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\author{Boris Ischi}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\address{Boris Ischi, Laboratoire de
Physique des Solides, Universit\'e Paris-Sud, B\^atiment 510,
91405 Orsay, France}
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\email{ischi@kalymnos.unige.ch}
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\thanks{Supported by the Swiss National Science Foundation.}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subjclass{Primary 06B23, 06C15; Secondary 81P10}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%\date{\today}
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\keywords{Complete atomistic ortholattice, tensor product, quantum
logic}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{abstract}
Given two complete atomistic lattices $\l_1$ and $\l_2$, we define
a set $\sep=\sep(\l_1,\l_2)$ of complete atomistic lattices by
means of three axioms (natural regarding the description of {\it
separated} quantum compound systems), or in terms of a universal
property with respect to a given class of bimorphisms. We prove
that $\sep$ is a complete lattice. The bottom element
$\l_1\aerts\l_2$ is the {\it separated product} of Aerts. For
atomistic lattices with $1$ (not complete),
$\l_1\aerts\l_2\cong\l_1\square\l_2$ the {\it box product} of
Gr\"atzer and Wehrung, and, in case $\l_1$ and $\l_2$ are moreover
coatomistic, $\l_1\aerts\l_2\cong\l_1\boxtimes\l_2$ the {\it
lattice tensor product}. The top element $\l_1\otf\l_2$ is the
(complete) join-semilattice tensor product of Fraser, which is
isomorphic to the tensor products of Chu and Shmuely. With some
additional hypotheses on $\l_1$ and $\l_2$ (true if $\l_1$ and
$\l_2$ are moreover orthomodular with the covering property), we
prove that $\sep$ is a singleton if and only if $\l_1$ or $\l_2$
is distributive, if and only if $\l_1\otf\l_2$ has the covering
property. Our main result reads: $\l\in\sep$ admits an
orthocomplementation if and only if $\l=\l_1\aerts\l_2$. For
$\l_1$ and $\l_2$ moreover irreducible, we characterize the
automorphisms of each $\l\in\sep$ in terms of those of $\l_1$ and
$\l_2$. At the end, we construct an example $\l_1\ots\l_2$ in
$\sep$ which has the covering property.
\end{abstract}
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\maketitle
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\section{Introduction}

In quantum logic, on associates to a physical system $S$ a couple
$(\s_S,\l_S\subseteq 2^{\s_S})$, where $\s_S$ represents all
possible states of $S$ and $\l_S$ the experimental propositions
concerning $S$: A proposition represented by some $a\in\l_S$ is
true with probability $1$ if and only if the state of $S$ lies in
$a$ \cite{Birkhoff/Neumann:1936}.

A cornerstone is the following theorem, by which the Hilbert space
structure of quantum mechanics can be recovered from a certain
number of axioms on the poset $\l_S$ \cite{Piron:1964} (see
\cite{Maeda/Maeda:handbook}, Theorems 34.5 and 34.9).

\begin{theorem}\label{TheoremPiron}
If $\l_S$ is an irreducible orthocomplemented simple closure space
on $\s_S$ (i.e. $\l_S$ is closed under arbitrary set-intersections
and contains $\emptyset$, $\s_S$ and all singletons) with the
covering property of length $\geq 4$, then there is a
$^*-$division ring $\mathbb{K}$ and a vector space $E_S$ over
$\mathbb{K}$ with an Hermitian form such that $\l_S$ is
ortho-isomorphic to the lattice $\pro(E_S)$ of $E_S-$closed
subspaces of $E_S$. Moreover, if $\mathbb{K}=\mathbb{R}$ or
$\mathbb{C}$ with the usual involution, then $E_S$ is a Hilbert
space if and only if $\l_S$ is also orthomodular.
\end{theorem}

\vspace{0.1cm}If $S$ is a compound system consisting of two
subsystems $S_1$ and $S_2$, then
$\s_{S_1}\t\s_{S_2}\subseteq\s_S$. In classical mechanics there is
equality whereas in quantum theory $\s_S$ is given by all
one-dimensional subspaces of the Hilbert space $\h_S=\h_{S_1}\ot
\h_{S_2}$ \cite{vonNeumann:handbook}. However, when two quantum
systems $S_1$ and $S_2$ are {\it separated} we have
\cite{Aerts:1982,IschiDenver:2004}
%
\begin{enumerate}
%
\item[(P1)] $\s_S=\s_{S_1}\t\s_{S_2}$.
%
\end{enumerate}
%
In particular, simultaneous experiments on both systems can be
performed and any experiment done on one system does not alter the
state of the other system. Therefore, if $P_1$ is a proposition
concerning $S_1$ represented by some $a_1\in\l_1$ and $P_2$ is a
proposition concerning $S_2$ represented by some $a_2\in\l_2$,
then the proposition $P_1$ OR $P_2$ concerning the compound system
is true with probability $1$ if and only if the state $p_1$ of
$S_1$ lies in $a_1$ or the state $p_2$ of $S_2$ lies in $a_2$. In
other words \cite{Aerts:1982,IschiDenver:2004}
%
\begin{enumerate}
%
\item[(P2)] $a_1\t\s_{S_2}\cup\s_{S_1}\t a_2\in\l_{S}$, for all
$a_1\in\l_{S_1}$ and $a_2\in\l_{S_2}$.
%
\end{enumerate}
%

For separated systems, it can be shown that $\l_S$, which we
denote by $\l_{S_{sep}}$, cannot be isomorphic to the lattice of
closed subspaces of a Hilbert space \cite{AertsHPA84}. As
consequence some axioms of Theorem \ref{TheoremPiron} are failing
in $\l_{S_{sep}}$.

In \cite{Aerts:1982}, Aerts proposed a model
$\l_{S_1}\aerts\l_{S_2}$ for $\l_{S_{sep}}$, call the {\it
separated product}. For $\l_{S_1}$ and $\l_{S_2}$
orthocomplemented simple closure spaces (in that case the
separated product is an orthocomplemented simple closure space),
he proved that if $\l_{S_1}\aerts\l_{S_2}$ has the covering
property or is orthomodular, then $\l_{S_1}$ or $\l_{S_2}$ is a
power set, hence $S_1$ or $S_2$ is a classical system
\cite{Aerts:1982}. As a consequence, according to Aerts, the
covering property and orthomodularity do not hold in
$\l_{S_{sep}}$. A similar conclusion concerning orthomodularity
was obtained by Pulmannov\'a in \cite{Pulmannova:1985}.

Here we show that $\l_{S_{sep}}$ does not admit an
orthocomplementation. Moreover we provide a natural model for
$\l_{S_{sep}}$ which has the covering property. We proceed as
follows: Following \cite{Piron:handbook} and \cite{Aerts:1982}, we
assume that $\l_{S_1}$, $\l_{S_2}$ and $\l_{S_{sep}}$ are simple
closure spaces. Moreover, in addition to Axioms P1 and P2 above,
we postulate that
%
\begin{enumerate}
%
\item[(P3)] for all $p_i\in\s_{S_i}$ and all
$A_i\subseteq\s_{S_i}$, [$p_1\t A_2\in\l_{S_{sep}}\Rightarrow
A_2\in\l_2$] and [$A_1\t p_2\in\l_{S_{sep}}\Rightarrow
A_1\in\l_1$].
%
\end{enumerate}
%
We refer to \cite{IschiDenver:2004} for detailed physical
justifications of Axioms P1-P3. Let
$\sep_{12}=\sep(\l_{S_1},\l_{S_2})$ be the set of all simple
closure spaces on $\s_{S_1}\t\s_{S_2}$ for which Axioms P2 and P3
hold. Let $\l\in\sep_{12}$. We prove that if $\l_{S_1}$ and
$\l_{S_2}$ are orthocomplemented simple closure spaces with the
covering property, then $\l$ admits and orthocomplementation if
and only if $\l=\l_{S_1}\aerts\l_{S_2}$. We conclude by a simple
physical argument which shows that certainly
$\l_{S_1}\aerts\l_{S_2}\subsetneqq\l_{S_{sep}}$
\cite{IschiDenver:2004}.

The rest of the paper is organized as follows. In Section
\ref{SectionMainDefinitions} we fixe some basic terminology and
notations and we define $\sep\equiv\sep(\l_1,\cdots,\l_n)$. We
prove that $\sep$ is a complete lattice, the bottom and top
elements of which are denoted by $\aerts_i\l_i$ and $\otf_i\l_i$
respectively. In Section \ref{SectionComparison} we compare
$\aerts$ and $\otf$ to a certain number of tensor products. An
equivalent definition of $\sep$, in terms of a universal property
with respect to a given class of bimorphisms (arbitrary joins are
preserved) is given in Section \ref{SectionEquivalentDefinition}.
Let $\l\in\sep$. In Section \ref{SectionCentralElements} we prove
some basic relations between the central elements of $\l_i$'s and
$\l$.

Further, with some additional hypotheses on each $\l_i$ (true if
$\l_i$ is moreover orthomodular with the covering property) we
prove that $\sep$ is a singleton ({\it i.e.}
$\aerts_i\l_i=\otf_i\l_i$) if and only if at most one $\l_i$ is
not a power set (Section
\ref{SectionSufficientandNecessaryConditionsFor...}) if and only
if $\aerts_i\l_i$ or $\otf_i\l_i$ has the covering property
(Section \ref{SectionCoevringProperty}). Recall that in the
physical interpretation, $\l_{S_i}=2^{\s_{S_i}}$ if and only if
$S_i$ is classical.

Let $\l\in\sep$ and $u:\l\rightarrow\l$ preserving arbitrary joins
and sending atoms to atoms. In Section \ref{SectionAutomorphisms}
we prove, under some hypotheses on the image of $u$, that if each
$\l_i$ is moreover irreducible, then there is a permutation $f$
and join-preserving maps $v_i:\l_i\rightarrow\l_{f(i)}$ sending
atoms to atoms, such that on atoms $u=f\circ (v_1\t\cdots\t v_n)$.
A time evolution can be modelled by a map preserving arbitrary
joins and sending atoms to atoms \cite{Faure/Moore/Piron:1995}.
Hence, in the physical interpretation, this result shows that
separated quantum systems remain separated only if they do not
interact. Finally, Section \ref{SectionOrthocomplementation} is
devoted to our main result and Section \ref{SectionAnExample} to
the example mentioned above.
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Main definitions}\label{SectionMainDefinitions}

In this section we give our main definitions. We start with some
background material and basic notations used in the sequel.

\begin{definition}
By a {\it simple closure space} $\l$ (on $\s$) we mean a set of
subsets of a nonempty set $\s$, ordered by set-inclusion, closed
under arbitrary set-intersections ({\it i.e.} for all
$\omega\subseteq \l$, $\cap\omega\in\l$), and containing $\s$,
$\emptyset$, and all singletons. We denote the bottom
($\emptyset$) and top ($\s$) elements by $0$ and $1$ respectively.
For $p\in\s$, we denote the singleton $\{p\}$ by $p$. Hence $p\cup
q$ stands for $\{p,q\}$.
\end{definition}

\begin{remark} Let $\l$ be a simple closure space on a (nonempty) set $\s$.
Then $\l$ is a complete atomistic lattice, the atoms of which
correspond to all singletons of $\s$. Note that if $A\subseteq\s$,
then $\vee_\l(A)=\cap\{b\in\l\,;\,A\subseteq b\}$.

Conversely, let $\l$ be a complete atomistic lattice. Let $\s$
denote the set of atoms of $\l$, and, for each $a\in\l$, let
$\s[a]$ denote the set of atoms under $a$. Then
$\{\s[a]\,;\,a\in\l\}$ is a simple closure space on $\s$,
isomorphic to $\l$.
\end{remark}

\begin{definition}
A lattice $\l$ with $0$ and $1$ is {\it orthocomplemented} if
there is a unary operation $\,^\p$ (orthocomplementation), also
denoted by $'$, such that for all $a,\,b\in\l$, $(a^\p)^\p=a$,
$a\leq b$ implies $b^\p\leq a^\p$, and $a\vee a^\p=1$.

An orthocomplemented lattice is {\it orthomodular} if for all
$a,\,b\in\l$, $a\leq b$ implies $b=a\vee(b\land a^\p)$.

A lattice with $0$ has the {\it covering property} if for any atom
$p$ and any $a\in\l$, $p\land a=0$ implies that $p\vee a$ covers
$a$ (in symbols $p\vee a\gtrdot a$).

Finally, a lattice $\l$ with $0$ and $1$ is called a {\it
DAC-lattice} if $\l$ and its dual $\l^*$ (defined by the converse
order-relation) are atomistic with the covering property
\cite{Maeda/Maeda:handbook}. We say that a lattice $\l$ with $1$
is {\it coatomistic} if the dual $\l^*$  is atomistic.
\end{definition}

\begin{remark}
Note that an orthocomplemented atomistic lattice with the covering
property is a DAC-lattice. Note also that in Theorem
\ref{TheoremPiron}, if instead of orthocomplemented, the simple
closure space is a DAC-lattice, then there is a pair of dual
vector spaces such that a representation theorem similar to
Theorem \ref{TheoremPiron} holds (see \cite{Maeda/Maeda:handbook},
Theorem 33.7).
\end{remark}

\begin{definition} We denote the category
of simple closure spaces with maps preserving arbitrary joins
(hence $0$) by $\cCal$, and the sub-category of simple closure
spaces on a particular (nonempty) set $\s$, by $\cCal(\s)$. $2$
denotes the simple closure space isomorphic to the two-element
lattice.

Let $\l\in\cCal(\s)$. We write $\aut(\l)$ for the group of
automorphisms of $\l$. Let $\l_1\in\cCal(\s_1)$. Note that any map
$u:\l\rightarrow\l_1$ sending atoms to atoms induces a mapping
from $\s$ to $\s_1$, which we also call $u$. Thus, if
$u\in\aut(\l)$, then for all $a\in\l$, $u(a)=\{u(p)\,;\,p\in a\}$.

If $\l$ is orthocomplemented, for $p,\, q\in\s$, we write $p\p q$
if and only if $p\in q^\p$, where $q^\p$ stands for $\{q\}^\p$.

Finally, if $\h$ is a complex Hilbert space, then $\s_\h$ denotes
the set of one-dimensional subspaces of $\h$ and $\pro(\h)$ stands
for the simple closure space isomorphic to the lattice of closed
subspaces of $\h$. Moreover, we write $\uni(\h)$ for the set of
automorphisms of $\pro(\h)$ induced by unitary maps on $\h$.
\end{definition}

\begin{definition}Let $\{\s_\alpha\}_{\alpha\in\Omega}$ be a
family of nonempty sets, $\mathbf{\s}=\prod_\alpha\s_\alpha$,
$\beta\in\Omega$, $p\in\mathbf{\s}$, and $R\subseteq\mathbf{\s}$.
We shall make use of the following notations:
%
\begin{enumerate}
%
\item We denote by $\pi_\beta:\mathbf{\s}\rightarrow\s_\beta$ the
$\beta-$th coordinate map, {\it i.e.}, $\pi_\beta(p)=p_\beta$.
%
\item We denote by $p[-,\beta]:\s_\beta\rightarrow\mathbf{\s}$ the
map which sends $q\in\s_\beta$ to the element of $\mathbf{\s}$
obtained by replacing $p$'s $\beta-$th entry by $q$.
%
\item We define $R_\beta[p]=\pi_\beta(p[\s_\beta,\beta]\cap R)$.
%
\end{enumerate}
We omit the $\beta$ in $p[-,\beta]$ when no confusion can occur.
For instance, we write $p[\s_\beta]$ instead of
$p[\s_\beta,\beta]$.
\end{definition}

\begin{remark}
$p[R_\beta[p]]=p[\s_\beta]\cap R$.
\end{remark}

\begin{definition}\label{DefinitionPTensorProduct}
Let $\{\l_\alpha\in\cCal(\s_\alpha)\}_{\alpha\in\Omega}$. We
denote by $\sep(\l_\alpha,\alpha\in\Omega)$ the set all
$\l\in\cCal(\mathbf{\s})$ such that
\begin{enumerate}
%
\item[(P1)] $\mathbf{\s}=\prod_\alpha\s_\alpha$,
%
\item[(P2)] $\cup_\alpha \pr_\alpha(a_\alpha)\in\l$, for all
$a\in\prod_\alpha\l_\alpha$,
%
\item[(P3)] for all $p\in\mathbf{\s}$, $\beta\in\Omega$, and
$A\subseteq\s_\beta$, [$p[A,\beta]\in\l\Rightarrow A\in\l_\beta$].
%
\end{enumerate}
%
Let $T=\prod_\alpha T_\alpha$ with $T_\alpha\subseteq
\aut(\l_\alpha)$. We denote by $\Sep_T(\l_\alpha,\alpha\in\Omega)$
the set of all $\l\in\sep(\l_\alpha,\alpha\in\Omega)$ such that
\begin{enumerate}
%
\item[(P4)] for all $ v\in T$, there is $u\in \aut(\l)$ such that
$u(p)_\alpha=v_\alpha(p_\alpha)$ for all $p\in\mathbf{\s}$ and
$\alpha\in\Omega$.
%
\end{enumerate}
We call elements of $\sep(\l_\alpha,\alpha\in\Omega)$ {\it weak
tensor products}.
\end{definition}

\begin{definition}
We say that $\l\in\cCal(\s)$ (respectively $T\subseteq \aut(\l)$)
is {\it transitive} if the action of $\aut(\l)$ (respectively the
action of $T$) on $\s$ is transitive.
\end{definition}

\begin{remark} The $u$ in Axiom P4 is necessarily unique.
If all $T_\alpha$ in Axiom P4 are transitive, then $\l$ is
transitive. Note also that,
$\Sep_T(\l_\alpha,\alpha\in\Omega)\subseteq\sep(\l_\alpha,\alpha\in\Omega)$
for all $T=\prod_\alpha T_\alpha$, with
$T_\alpha\subseteq\aut(\l_\alpha)$. The name ``weak tensor
product'' is justified by the fact that
$\sep(\l_\alpha,\alpha\in\Omega)$ and
$\Sep_T(\l_\alpha,\alpha\in\Omega)$ can be defined in terms of a
universal property with respect to a given class of bimorphisms of
$\cCal$ (see Section \ref{SectionEquivalentDefinition}).
\end{remark}

\begin{definition}\label{DefinitionSeparatedProduct}
Let $\{\s_\alpha\}_{\alpha\in\Omega}$ be a family of nonempty sets
and $\{\l_\alpha\subseteq 2^{\s_\alpha}\}_{\alpha\in\Omega}$. Let
$\mathbf{\s}=\prod_\alpha\s_\alpha$. Then
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\[\begin{split}\aerts\{\l_\alpha\,;\,\alpha\in\Omega\}&:=\{\cap \omega\,;\,
\omega\subseteq \{\cup_\alpha \pr_\alpha(a_\alpha)\,
;\, a\in\mbox{$\prod$}_\alpha\l_\alpha\}\}\, ,\\
%
\otf\{\l_\alpha\,;\,\alpha\in\Omega\}&:=\{R\subseteq\mathbf{\s}\,;\,
R_\beta[p]\in\l_\beta,\, \forall
p\in\mathbf{\s},\,\beta\in\Omega\}\, ,
\end{split}\]
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
ordered by set-inclusion.
\end{definition}

\begin{proposition}\label{PropositionAssociativity}
Let $\{\s_\alpha\}_{\alpha\in\Omega}$ be a family of nonempty
sets, and let $\{\l_\alpha\subseteq
2^{\s_\alpha}\}_{\alpha\in\Omega}$. Let
$\{\Omega_\gamma\subseteq\Omega\,;\,\gamma\in\Gamma\}$ such that
$\Omega=\coprod\{\Omega_\gamma\,;\,\gamma\in\Gamma\}$. Then
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\[\begin{split}
\aerts\{\l_\alpha\,;\,\alpha\in\Omega\}&=
\aerts_{\gamma\in\Gamma}(\aerts_{\alpha\in\Omega_\gamma}
\l_\alpha)\, ,\\
%
\otf\{\l_\alpha\,;\,\alpha\in\Omega\}&=
\otf_{\gamma\in\Gamma}(\otf_{\alpha\in\Omega_\gamma} \l_\alpha)\,
.
\end{split}
\]
\end{proposition}

\begin{proof} Direct from Definition
\ref{DefinitionSeparatedProduct}. \end{proof}

\begin{theorem}\label{TheoremTopandBottomElements}
Let $\{\l_\alpha\in\cCal(\s_\alpha)\}_{\alpha\in\Omega}$. For all
$T=\prod_\alpha T_\alpha$ with $T_\alpha\subseteq\aut(\l_\alpha)$,
$\aerts_\alpha\l_\alpha$ and $\otf_\alpha\l_\alpha$ are in
$\Sep_T(\l_\alpha,\alpha\in\Omega)$. Moreover,
$\aerts_\alpha\l_\alpha$ and $\otf_\alpha\l_\alpha$ are the bottom
and the top elements of $\sep(\l_\alpha,\alpha\in\Omega)$
respectively, and $\Sep_T(\l_\alpha,\alpha\in\Omega)$ and
$\sep(\l_\alpha,\alpha\in\Omega)$, ordered by set-inclusion, are
complete lattices.
\end{theorem}

\begin{proof} Let $\mathbf{\s}=\prod_\alpha\s_\alpha$.
Obviously, $\aerts_\alpha\l_\alpha$ and $\otf_\alpha\l_\alpha$
contain $\emptyset$ and $\mathbf{\s}$, and by definition
$\aerts_\alpha\l_\alpha$ is $\cap-$closed and
$\otf_\alpha\l_\alpha$ contains all singletons. If
$\omega\subseteq\otf_\alpha\l_\alpha$, then
$(\cap\omega)_\beta[p]=\cap\{R_\beta[p]\,;\, R\in\omega\}$.
Moreover, for all $p\in\mathbf{\s}$, $\cap_\alpha
\pr_\alpha(p_\alpha)=p$. As a consequence,
$\aerts_\alpha\l_\alpha$ and $\otf_\alpha\l_\alpha$ are simple
closure spaces on $\mathbf{\s}$, and obviously, Axioms P2 and P3
hold (note that for $A\subseteq\s_\beta$,
$p[A,\beta]\subseteq\cup_\alpha \pr_\alpha(a_\alpha)$ if and only
if $A\subseteq a_\beta$, or $p_\gamma\in a_\gamma$ for some
$\gamma\ne\beta$, therefore $\vee_\aertssub p[A,\beta]=p[\vee
A,\beta]$).

Let $v\in T$, $a\in\prod_\alpha\l_\alpha$, $p\in\mathbf{\s}$,
$R\subseteq\mathbf{\s}$, and $\omega\subseteq 2^{\mathbf{\s}}$.
Then,
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\[v(\cup_\alpha \pr_\alpha(a_\alpha))=\cup_\alpha
\pr_\alpha(v_\alpha(a_\alpha))\,
,\]
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
and
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\[v(R)_\beta[v(p)]=v_\beta(R_\beta[p])\, .\]
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
Moreover $v(\cap\omega)=\cap\{v(x)\,;\,x\in\omega\}$. Hence, $v$
is a bijection of $\aerts_\alpha\l_\alpha$ and of
$\otf_\alpha\l_\alpha$, and $v$ preserves arbitrary meets, hence
also arbitrary joins.

Let $\l\in\sep(\l_\alpha,\alpha\in\Omega)$. By Axiom P2,
$\aerts_\alpha\l_\alpha\subseteq\l$, whereas by Axiom P3,
$\l\subseteq\otf_\alpha\l_\alpha$.

Finally, if $\omega\subseteq\sep(\l_\alpha,\alpha\in\Omega)$
(respectively $\omega\subseteq\Sep_T(\l_\alpha,\alpha\in\Omega)$),
then obviously $\cap\omega=\{a\in\l\,,\,\forall
\l\in\omega\}\in\sep(\l_\alpha,\alpha\in\Omega)$ (respectively
$\cap\omega\in\Sep_T(\l_\alpha,\alpha\in\Omega)$).\end{proof}

\begin{corollary} Let
$\{\l_\alpha\in\cCal(\s_\alpha)\}_{\alpha\in\Omega}$ and
$\mathbf{\s}=\prod_\alpha\s_\alpha$. Then
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\[\sep(\l_\alpha,\alpha\in\Omega)=\{\l\in\cCal(\mathbf{\s})\,;\,
\aerts_\alpha\l_\alpha\subseteq\l\subseteq\otf_\alpha\l_\alpha\}\,
.\]
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\end{corollary}

\begin{remark} Let $\{\l_\alpha\in\cCal\}_{\alpha\in\Omega}$. For
all $T=\prod_\alpha T_\alpha$ with
$T_\alpha\subseteq\aut(\l_\alpha)$,
$\Sep_T(\l_\alpha,\alpha\in\Omega)$ is a complete meet-sublattice
of $\sep(\l_\alpha,\alpha\in\Omega)$.\end{remark}

\begin{proposition} Let $\l_1,\,\l_2\in\cCal$. Then
$\l_1\aerts\l_2\cong\l_2\aerts\l_1$,
$\l_1\otf\l_2\cong\l_2\otf\l_1$, and $2\aerts\l_1\cong\l_1\cong
2\otf\l_1$.\end{proposition}

\begin{proof} Direct from Definition
\ref{DefinitionSeparatedProduct}.\end{proof}

\begin{proposition} Let $\{\l_\alpha\in\cCal\}_{\alpha\in\Omega}$
and $T=\prod_\alpha T_\alpha$ with
$T_\alpha\subseteq\aut(\l_\alpha)$. Then there is an isomorphism
$f:\Sep_T(\l_\alpha,\alpha\in\Omega)\rightarrow
\Sep_T(2,\l_\alpha,\alpha\in\Omega)$ such that for all
$\l\in\Sep_T(\l_\alpha,\alpha\in\Omega)$, $\l\cong f(\l)$.
\end{proposition}

\begin{proof} Direct from Definition
\ref{DefinitionPTensorProduct}.\end{proof}

\begin{example}
We now consider two examples, well known in many-body quantum
physics, where instead of Axioms P1 and P2, we have
\begin{enumerate}
%
\item[(p1)] $\prod_\alpha\s_\alpha\subseteq\mathbf{\s}$,
%
\item[(p2)] $(\prod_\alpha\s_\alpha)\cap(\vee(\cup_\alpha
\pr_\alpha(a_\alpha)))=\cup_\alpha \pr_\alpha(a_\alpha)$, $\forall
a\in\prod_\alpha\l_\alpha$,
%
\end{enumerate}

Let $\h_1$ and $\h_2$ be complex Hilbert spaces,
$\l_1=\pro(\h_1)$, $\l_2=\pro(\h_2)$ and $\l=\pro(\h_1\ot\h_2)$.
Then, Axioms p1, p2 and P3 hold in $\l$. Moreover, Axiom P4 holds
for $T=\uni(\h_1)\t\uni(\h_2)$. Note that $\vee (A_1\t A_2)=1$,
for all $A_i\subseteq\s_{\h_i}$ with $\vee A_i=1$.

Let $\h$ be a complex Hilbert space and
$\mathcal{F}=\bigoplus_{n\geq 0} \h^{\ot n}$ be the Fock space
(neither symmetrized, nor antisymmetri\-zed). Let
$\l=\pro(\mathcal{F})$, and for all integer $i$, let
$\l_i=\pro(\h)$. Let $n$ be an integer. Consider the family
$\{\l_i\,;\,1\leq i\leq n\}$. Then Axioms p1, p2, P3 and P4 with
$T_i=\uni(\h)$ hold in $\l$ for all $n$. For all $i$, let
$\s_i=\s_\h$. Note that in that case, for all $n$ we have
$\vee(\prod_{i=1}^n\s_i)\ne 1$.
\end{example}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Comparison with other tensor
products}\label{SectionComparison}

In this section, we compare the bottom element $\l_1\aerts\l_2$
with the {\it separated product} of Aerts, with the {\it box
product} $\l_1\square\l_2$ of Gr\"atzer and Wehrung, and with the
{\it lattice tensor product} $\l_1\boxtimes\l_2$. On the other
hand, we compare the top element $\l_1\otf\l_2$ with the
semilattice tensor product of Fraser and with the tensor products
of Chu, Golfin and Shmuely.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{The separated product}

\begin{definition}[D. Aerts, \cite{Aerts:1982}]\label{DefinitionOrthAerts}
Let $\{\l_\alpha\in\cCal(\s_\alpha)\}_{\alpha\in\Omega}$ with for
all $\alpha\in\Omega$, $\l_\alpha$ orthocomplemented,
$\mathbf{\s}=\prod_\alpha\s_\alpha$, and let
$p,\,q\in\mathbf{\s}$. Denote by $\#$ the binary relation on
$\mathbf{\s}$ defined by $p\# q$ if and only if there is
$\beta\in\Omega$ such that $p_\beta\p_\beta q_\beta$. Then
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\[\otaerts_\alpha\l_\alpha:=\{ R\subseteq \mathbf{\s};\
R^{\#\#}=R\}\, .\]
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\end{definition}

\begin{lemma}
Let $\{\l_\alpha\in\cCal(\s_\alpha)\}_{\alpha\in\Omega}$ with for
all $\alpha\in\Omega$, $\l_\alpha$ orthocomplemented. Then
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\[\otaerts_\alpha\l_\alpha=\aerts_\alpha\l_\alpha\, ,\]
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
and $\otaerts_\alpha\l_\alpha$ is orthocomplemented.\end{lemma}

\begin{proof}
Let $\mathbf{\s}=\prod_\alpha\s_\alpha$. Obviously, $\#$ is
symmetric and anti-reflexive. Since $\l_\alpha$ is
orthocomplemented, $\p_\alpha$ is {\it separating}, {\it i.e.} for
all $p_\alpha,$ $q_\alpha\in\s_\alpha$, there is
$r_\alpha\in\s_\alpha$ such that $p_\alpha\p_\alpha r_\alpha$ and
$q_\alpha\not\p_\alpha r_\alpha$. Therefore, it follows directly
from Definition \ref{DefinitionOrthAerts} that $\#$ is separating.
As a consequence, $\otaerts_\alpha\l_\alpha$ is an
orthocomplemented simple closure space on $\mathbf{\s}$. Moreover,
coatoms are given by $p^\#=\cup_\alpha
\pr_\alpha(p_\alpha^{\p_\alpha})$. As a consequence,
$\otaerts_\alpha\l_\alpha\subseteq\aerts_\alpha\l_\alpha$.

Let $a\in\prod_\alpha\l_\alpha$. Denote the set of coatoms above
$a_\alpha$ by $\s'[a_\alpha]$. Then
$a_\alpha=\cap_\alpha\s'[a_\alpha]$. Moreover,
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\[\bigcap\{\cup_\alpha
\pr_\alpha(x_\alpha)\,;\,x\in\mbox{$\prod$}_\alpha\s'[a_\alpha]\}=
\cup_\alpha
\pr_\alpha(a_\alpha)\, .\]
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
As a consequence,
$\aerts_\alpha\l_\alpha\subseteq\otaerts_\alpha\l_\alpha$.
\end{proof}

\begin{remark} The symbol $\aerts$ was originally used by
Aerts.\end{remark}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{The box product and the lattice tensor product}

\begin{definition}[G. Gr\"atzer, F. Wehrung,
\cite{Graetzer/Wehrung:1999}]\label{DefinitionWehrung} Let $\l_1$
and $\l_2$ be lattices and $(a,b)\in\l_1\t\l_2$. Define $a\square
b=\{(x,y)\in\l_1\t\l_2\,;\, x\leq a$ or $y\leq b\}$. The {\it box
product} $\l_1\square\l_2$ is defined as the set of all finite
intersections of the form $\cap \{a_i\square b_i\,;\, i\leq n\}$
where $(a_i,b_i)\in\l_1\t\l_2$ for all $i\leq n$, ordered by
set-inclusion (\cite{Graetzer/Wehrung:1999}, Definition 2.1).

Define $a\boxtimes b=\{(x,y)\in\l_1\t\l_2\,;\, x\leq a$ and $y\leq
b\}\cup (\l_1\t\bot_{\l_2})\cup(\bot_{\l_1}\t\l_2)$ where
$\bot_{\l_i}=\{0_i\}$ if $\l_i$ has a zero, and $\emptyset$
otherwise. Call a subset $X$ of $\l_1\t\l_2$ {\it confined} if
$X\subseteq a\boxtimes b$ for some $(a,b)\in\l_1\t\l_2$. Let
$\l_1\boxtimes\l_2$ be the set of all confined elements of
$\l_1\square\l_2$, ordered by set-inclusion. If
$\l_1\boxtimes\l_2$ is nonempty, then it is called the {\it the
lattice tensor product} of $\l_1$ and $\l_2$ (see
\cite{Graetzer/Wehrung:1999}, Section 3).
\end{definition}

\begin{definition}\label{DefinitionAertsFinite}
Let $\l$ be a lattice and $a\in\l$. We denote by $a\downarrow$ the
set $\{x\in\l\,;\, x\leq a\}$ and by $L$ the set
$\{a\downarrow\,;\, a\in\l\}$ ordered by set-inclusion (note that
$L\subseteq 2^{\l}$).

Let $\s_1$ and $\s_2$ be nonempty sets, $\l_1\subseteq 2^{\s_1}$
and $\l_2\subseteq 2^{\s_2}$. We define $\l_1\aerts_n\l_2$ by
taking only finite intersections in Definition
\ref{DefinitionSeparatedProduct}. If $\l_1$ and $\l_2$ are
atomistic lattices, then $\l_1\aerts_n\l_2$ is defined as
$l_1\aerts_n l_2$, where $l_i=\{\s[a_i]\,;\,a_i\in\l_i\}\subseteq
2^{\s_i}$ and $\s[a_i]$ denotes the set of atoms under $a_i$
($\s_i:=\s[1_i]$).
\end{definition}

\begin{proposition}
Let $\l_1$ and $\l_2$ be lattices with $1$. Then
$\l_1\square\l_2=L_1\aerts_n L_2$.\end{proposition}

\begin{proof} Direct from Definitions \ref{DefinitionWehrung} and
\ref{DefinitionAertsFinite}.\end{proof}

\begin{lemma}[\cite{Graetzer/Wehrung:1999}, Corollary
4.2]\label{LemmaWehrung} Let $\l_1$ and $\l_2$ be  lattices with
$0$. Then $\l_1^*\square\l_2^*\cong(\l_1\boxtimes\l_2)^*$.
\end{lemma}

\begin{theorem}
For atomistic lattices with $1$ (and $0$),
$\l_1\aerts_{n}\l_2\cong\l_1\square\l_2$, and in case $\l_1$ and
$\l_2$ are moreover coatomistic,
$\l_1\aerts_n\l_2\cong\l_1\boxtimes\l_2$.
\end{theorem}

\begin{proof} Define $f:\l_1\aerts_n\l_2\rightarrow\l_1\square\l_2$ as
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\[f(\s[a_1]\t \s_2\cup \s_1\t \s[a_2])=a_1\square
a_2:=(a_1\downarrow\nobreak)\t \l_2\cup \l_1\t (a_2\downarrow)\,
,\]
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
and $f(\cap x_i):=\cap f(x_i)$. Obviously, $f$ is bijective and
preserves meets, hence also joins.

From Lemma \ref{LemmaWehrung} and what precedes, we have
$(\l_1\boxtimes\l_2)^*\cong \l_1^*\aerts_n\l_2^*$. Now, an atom of
$\l_1^*\aerts_n\l_2^*$ is of the form $r=\{(x_1,x_2)\}$ with $x_1$
and $x_2$ coatoms, and a coatom is of the form
$x=\s'[p_1]\t\s_2'\cup\s_1'\t\s'[p_2]$ with $p_1$ and $p_2$ atoms
and where $\s'[a]$ denotes the set of coatoms above $a$
($\s_i':=\s'[0_i]$). Moreover, $r\leq x$ if and only if $p_1\leq
x_1$ or $p_2\leq x_2$.

On the other hand, an atom of $(\l_1\aerts_n\l_2)^*$ is of the
form $s=\s[x_1]\t\s_2\cup\s_1\cup\s[x_2]$ and a coatom is of the
form $y=\{(p_1,p_2)\}$, and $s\leq y$ if and only if $p_1\leq x_1$
or $p_2\leq x_2$. Therefore, there is obviously an isomorphism
between $\l_1^*\aerts_n\l_2^*$ and $(\l_1\aerts_n\l_2)^*$, hence
between $\l_1\boxtimes\l_2$ and $\l_1\aerts_n\l_2$.\end{proof}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{The tensor products of Chu, Golfin and Shmuely}

\begin{definition}[P. H. Chu, \cite{Barr:handbook} \S II.1, III.3]
The category $\Chu(\cSet,2)$ has as objects triples $\A=(A,r,X)$
where $A$ and $X$ are sets and $r$ is a map from $A\t X$ to $2$
(where $2=\{0,1\}$), and as arrows pairs of maps
$(F,G):\A\rightarrow(B,s,Y)$ with $F:A\rightarrow B$ and
$G:Y\rightarrow X$ such that $s(F(a),y)=r(a,G(y))$ for all $a\in
A$ and $y\in Y$. There is a functor $\,^\p$ and a bifunctor
$\otchu$ defined on objects as $\A^\p=(X,\check{r},A)$ with
$\check{r}(x,a)=r(a,x)$, and $\A\otchu \B=(A\t
B,t,\Chu(\A,\B^\p))$ with
$t((a,b),(F,G))=r(a,G(b))=\check{s}(F(a),b)$. \end{definition}

\begin{remark} A simple closure space $\l\in\cCal(\s)$
is of course a Chu space, which we also denote by $\l$ (for
$p\in\s$ and $a\in\l$, put $r(p,a)=1$ if and only if $p\in
a$).\end{remark}

\begin{proposition}
Let $\l_0\in\cCal(\s_0)$ and $\l_1\in\cCal(\s_1)$. Then there is a
one to one correspondence between invertible arrows in
$\Chu(\l_0,\l_1)$ and isomorphisms between $\l_0$ and $\l_1$.
\end{proposition}

\begin{proof}
Let $f:\l_0\rightarrow\l_1$ be an isomorphism. Then
$(f,f^{-1})\in\Chu(\l_0,\l_1)$. Let $(F,G)$ be an invertible arrow
in $\Chu(\l_0,\l_1)$. Then $G:\l_1\rightarrow\l_0$ is bijective,
and since $G=F^{-1}$, $G$ preserves arbitrary meets, hence also
arbitrary joins.\end{proof}

\begin{definition}Let $\l_1$ and $\l_2$ be posets. A Galois connection
between $\l_1$ and $\l_2$ is a pair $(f,g)$ of order-preserving
maps with $f:\l_1\rightarrow\l_2$ and $g:\l_2\rightarrow\l_1$,
such that for all $a\in\l_1$ and $b\in\l_2$, $f(a)\leq
b\Leftrightarrow a\leq g(b)$.
\end{definition}

\begin{lemma}\label{LemmaJoin-preserving}
Let $\l_0$ and $\l_1$ be complete join-semilattices and $(f,g)$ a
pair of maps with $f:\l_0\rightarrow\l_1$ and
$g:\l_1\rightarrow\l_0$.
%
\begin{enumerate}
%
\item If $(f,g)$ forms a Galois connection, then $f$ preserves
arbitrary joins.

\item Suppose that $\l_0\in\cCal(\s_0)$. For $b\in\l_1$, let
$F^{-1}(b)$ denote the set $\{p\in\s_0\,;\,f(p)\leq b\}$. Then $f$
preserves arbitrary joins $\Leftrightarrow$ $f(a)=\vee
\{f(p)\,;\,p\in a\}$ for all $a\in\l_0$ and $F^{-1}(b)\in\l_0$ for
all $b\in\l_1$.
%
\end{enumerate}
\end{lemma}

\begin{proof} (1) Let $\omega\subseteq\l_0$. Since $f$ preserves
order, $y:=\vee\{f(x)\,;\,x\in\omega\}\leq f(\vee\omega)$. On the
other hand, for any $x\in\omega$, $f(x)\leq y$, hence $x\leq
g(y)$. As a consequence, $\vee\omega\leq g(y)$, therefore
$f(\vee\omega)\leq y$.

(2) ($\Rightarrow$) If $p\in\vee F^{-1}(b)$, then $f(p)\leq \vee
\{f(q)\,;\, q\in F^{-1}(b)\}\leq b$, hence $p\in F^{-1}(b)$.
($\Leftarrow$) Obviously, $(f,F^{-1})$ forms a Galois connection
between $\l_0$ and $\l_1$.
\end{proof}

\begin{theorem}\label{TheoremOtf=Chu}
Let $\l_1\in\cCal(\s_1)$ and $\l_2\in\cCal(\s_2)$. Then,
%
\begin{enumerate}
\item $\l_1\otf\l_2\cong\l_1\otchu\l_2$,
%
\item $(\l_1\otf\l_2)^*$ $\cong\{f:\l_1\rightarrow\l_2^*\,;\, f$
preserves arbitrary joins$\}$ ordered pointwise ({\it i.e.} $f\leq
g$ if and only if $f(p_1)\leq_* g(p_1)$ for all $p_1\in\s_1$,
where $\leq_*$ denotes the order relation in $\l_2^*$).
%
\end{enumerate}
%
\end{theorem}

\begin{proof} (1) We denote $\s_1\t\s_2$ by $\mathbf{\s}$.
Let $(F,G)\in\Chu(\l_1,\l_2^\p)$. Define
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\[R_{(F,G)}:=\bigcup_{p_1\in\s_1}p_1\t F(p_1)\, .\]
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
By definition, ${R_{(F,G)}}_2[p]\in\l_2$ for all
$p\in\mathbf{\s}$, and
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\[q_1\in {R_{(F,G)}}_1[p]\Leftrightarrow(q_1,p_2)\in
R_{(F,G)}\Leftrightarrow p_2\in F(q_1)\Leftrightarrow q_1\in
G(p_2)\, ,\]
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
thus ${R_{(F,G)}}_1[p]\in\l_1$ for all $p\in\mathbf{\s}$, hence
$R_{(F,G)}\in\l_1\otf\l_2$.

Let $R\in\l_1\otf\l_2$. Define $F_R:\s_1\rightarrow\l_2$ and
$G_R:\s_2\rightarrow\l_1$ as $F_R(p_1):=R_2[(p_1,\cdot)]$ and
$G_R(p_2):=R_1[(\cdot,p_2)]$. Then
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\[p_2\in F_R(p_1)\Leftrightarrow (p_1,p_2)\in R\Leftrightarrow
p_1\in G_R(p_2)\, ,\]
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
so that $(F_R,G_R)\in\Chu(\l_1,\l_2^\p)$.

Obviously,
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\[F_{R_{(F,G)}}=F,\ G_{R_{(F,F)}}=G,\ \mbox{and}\
R_{(F_R,G_R)}=R\, .\]
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
As a consequence, there is an invertible arrow in
$\Chu(\l_1\otf\l_2,\l_1\otchu\l_2)$.

(2) Let $(F,G)\in\Chu(\l_1,\l_2^\p)$. Define
$f:\l_1\rightarrow\l_2^*$ and $g:\l_2^*\rightarrow\l_1$ as
$f(a):=\cap F(a)$ and $g(b)=\cap G(b)$. Then, obviously $(f,g)$
forms a Galois connection between $\l_1$ and $\l_2^*$.

Let $f:\l_1\rightarrow\l_2^*$ be $\vee-$preserving. Define
$g:\l_2\rightarrow\l_1$ as $g(b):=\vee\{a\in\l_1\,; f(a)\leq_*
b\}$. Then, obviously $(f,g)\in\Chu(\l_1,\l_2^\p)$. \end{proof}

\begin{remark} Note that $\otf$ is the $\boxtimes$-tensor
product  of Golfin
(\cite{Golfin:handbook}, Definition II.3.6, II.1.1). As a
corollary of Theorem \ref{TheoremOtf=Chu} part 2,
$\l_1\otf\l_2\cong \l_1\ot\l_2$ the tensor product of Shmuely
\cite{Shmuely:1974}. Note that there is a one to one
correspondence between maps from $\l_1$ to $\l_2^*$ preserving
arbitrary joins and Galois connections between $\l_1$ and
$\l_2^*$.
\end{remark}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{The semilattice tensor product of Fraser}

\begin{definition} Let $\cC$ be a concrete
category over $\cSet$. Let $\{\A_\alpha\}_{\alpha\in\Omega}$ and
$\C$ be objects of $\cC$. A {\it multi-morphism} of $\cC$ is an
arrow $f\in\cSet(\prod_\alpha\A_\alpha,\C)$ such that for all
$a\in\prod_\alpha\A_\alpha$, $f(a[-,\alpha])\in\cC(\A_\alpha,\C)$,
where for $x\in\A_\alpha$, $a[x,\alpha]$ is defined as
$\pi_\alpha(a[x,\alpha])=x$ and $\pi_\beta(a[x,\alpha])=a_\beta$,
for all $\beta\ne\alpha$. If $\abs{\Omega}=2$, then $f$ is called
a {\it bimorphism}.
\end{definition}

\begin{definition}[G. Seal, \cite{Seal:2004}]\label{DefinitionTensorGavin}
Let $\cC$ be a concrete category over $\cSet$. A {\it tensor
product} in ${\bf C}$ is a bifunctor
$-\ot-:\cC\t\cC\rightarrow\cC$ such that for all objects $\A$,
$\B$ of $\cC$, there is a a bimorphism $f:
\A\t\B\rightarrow\A\ot\B$, and for all object $\C$ of $\cC$, and
all bimorphism $g:\A\t\B\rightarrow\C$, there is a unique arrow
$h\in\cC(\A\ot\B,\C)$ which makes the diagram
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
$$\bfig
 \ptriangle|alm|/>`>`>/[\A\t\B`\A\ot\B`\C;f`g`!h]
 \efig$$
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
commute.
\end{definition}

\begin{remark}
By definition, the tensor product is unique up to isomorphisms.
For the category of join-semilattices with maps preserving finite
joins, Definition \ref{DefinitionTensorGavin} is equivalent to the
definition of join-semilattice tensor product given by Fraser in
\cite{Fraser:1976} (note that $f(\l_1\t\l_2)$ generates
$\l_1\ot\l_2$ if and only if the arrow $h$ is unique). Therefore,
we call the tensor product in the category of complete
join-semilattices with maps preserving arbitrary joins, the
complete join-semilattice tensor product. For
$\l_1,\,\l_2\in\cCal$, it is given by $\l_1\otf\l_2$ (Theorem
\ref{TheoremTensorF=bi-tensor}).
\end{remark}

\begin{lemma}\label{LemmafisaBimorphism}
Let $\{\l_\alpha\in\cCal(\s_\alpha)\}_{\alpha\in\Omega}$, and
$\l\in\sep(\l_\alpha,\alpha\in\Omega)$. Then, for all
$a\in\prod_\alpha\l_\alpha$, $\prod_\alpha a_\alpha\in\l$.
Moreover, the map $f:\prod_\alpha\l_\alpha\rightarrow\l$ defined
as $f(a)=\prod_\alpha a_\alpha$, is an injective multi-morphism of
$\cCal$.
\end{lemma}

\begin{proof} Obviously $\prod_\alpha a_\alpha
=\cap_\alpha\pr(a_\alpha)$, hence the first statement follows from
Axiom P2.

Let $\mathbf{\s}=\prod_\alpha\s_\alpha$, $p\in\mathbf{\s}$,
$\beta\in\Omega$, $\omega\subseteq\l_\beta$, and
$a\in\prod_\alpha\l_\alpha$. By Axiom P3,
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\[\bigvee_{x\in\omega} p[x,\beta]=p[\vee\omega,\beta]\, .\]
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
As a consequence,
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\[\bigvee_{x\in\Omega}f(a[x,\beta])=
\bigvee_{x\in\omega}\bigvee_{p\in\prod_\alpha
a_\alpha}p[x,\beta]=\bigvee_{p\in\prod_\alpha
a_\alpha}p[\vee\omega,\beta]=f(a[\vee\omega,\beta])\, .\]
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\end{proof}

\begin{theorem}\label{TheoremTensorF=bi-tensor}
$\otf$  is a tensor product in $\cCal$.
\end{theorem}

\begin{proof} Let $\l_1,\,\l_2\in\cCal$. By Theorem
\ref{TheoremTopandBottomElements}, $\l_1\otf\l_2\in\cCal$. By
Lemma \ref{LemmafisaBimorphism}, the map $f:\l_1\t\l_2\rightarrow
\l_1\otf\l_2$ defined as $f(a_1,a_2):=a_1\t a_2$ is a bimorphism.

Let $\l\in\cCal$ (or a complete join-semilattice), and let
$g:\l_1\t\l_2\rightarrow \l$ be a bimorphism. Define
$h:\l_1\otf\l_2\rightarrow \l$ by $h(R)=\vee\{g(p_1,p_2)\,;\,
(p_1,p_2)\in R\}$. Let $b\in\l$. Then (see Lemma
\ref{LemmaJoin-preserving} for notations),
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\[ H^{-1}(b)=\bigcup_{p_1\in\s_1}p_1\t
G_{p_1}^{-1}(b)=\bigcup_{p_2\in\s_2}G_{p_2}^{-1}(b)\t p_2\, ,\]
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
where $g_{p_1}:=g(p_1,-)$ and $g_{p_2}:=g(-,p_2)$. Since $g$ is a
bimorphism, it follows from Lemma \ref{LemmaJoin-preserving} and
Definition \ref{DefinitionSeparatedProduct} that
$H^{-1}(b)\in\l_1\otf\l_2$. Therefore, by Lemma
\ref{LemmaJoin-preserving}, $h\in \cCal(\l_1\otf\l_2,\l)$.
Finally, if $h'\in\cCal(\l_1\otf\l_2,\l)$ and $h'\circ f=g$, then
$h'$ equals $h$ on atoms, therefore $h'=h$.

Let $L_1,\,L_2\in\cCal$, $f_1\in\cCal(\l_1,L_1)$, and
$f_2\in\cCal(\l_2,L_2)$. Then $g=f\circ(f_1\t
f_2):\l_1\t\l_2\rightarrow L_1\otf L_2$ is a bimorphism. We define
$f_1\otf f_2\in\cCal(\l_1\otf\l_2,L_1\otf L_2)$ to be the arrow
$h$ constructed above.
\end{proof}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Equivalent definition}\label{SectionEquivalentDefinition}

In this section we give an equivalent definition of
$\sep(\l_\alpha,\alpha\in\Omega)$ and
$\Sep_T(\l_\alpha,\alpha\in\Omega)$ in terms of a universal
property with respect to a given class of bimorphisms of $\cCal$.

\begin{lemma}\label{LemmaExitsh_a:L->2}
Let $\l\in\cCal(\s)$. For all $a\in\l$, there is (a unique)
$h_a:\l\rightarrow 2$ preserving arbitrary joins such that for all
$p\in\s$, $h_a(p)=0$ if and only if $p\in a$.
\end{lemma}

\begin{proof} Let $b\in\l$.
Define $h_a$ by $h_a(b)=\vee\{ h_a(p)\,;\, p\in b\}$ (hence
$h_a(0)=0$) with $h_a(p)=0$ if $p\in a$ and $h_a(p)=1$ if
$p\not\in a$. Then, by Lemma \ref{LemmaJoin-preserving}, $h$
preserves arbitrary joins.\end{proof}

\begin{lemma} Let $\l\in\cCal$, $\{\l_\alpha\in\cCal\}_{\alpha\in\Omega}$,
$T=\prod_\alpha T_\alpha$ with $T_\alpha\subseteq\aut(\l_\alpha)$,
and $f:\prod_\alpha\l_\alpha\rightarrow \l$ a multi-morphism of
$\cCal$. Define
%
\[\begin{split}
V_T(f)&:=\{f\circ v:\prod_\alpha\l_\alpha\rightarrow\l\,;\,v\in
T\}\,
,\\
%
\Gamma_\cap&:=\{\cap_\alpha
g_\alpha:\prod_\alpha\l_\alpha\rightarrow
2\,;\,g_\alpha\in\cCal(\l_\alpha,2)\}\, ,\end{split}
\]
%
with $\cap_\alpha g_\alpha(a):=\cap\{
g_\alpha(a_\alpha)\,;\,\alpha\in\Omega\}$. Then the maps in
$V_T(f)$ and $\Gamma_\cap$ are multi-morphisms of $\cCal$.
\end{lemma}

\begin{proof}
Let $g=\cap_\alpha g_\alpha\in\Gamma_\cap$,
$a\in\prod_\alpha\l_\alpha$, $\beta\in\Omega$, and $b\in\l$. Write
$h(-):=g(a[-,\beta]):\l_\beta\rightarrow 2$. Then,
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\[H^{-1}(0):=\{p_\beta\in\s_\beta\,;\,
g_\beta(p_\beta)\bigcap_{\alpha\ne\beta}g_\alpha(a_\alpha)=0\}\]
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
equals $1$ if $g_\alpha(a_\alpha)=0$ for some $\alpha\ne\beta$,
and $G_\beta^{-1}(0)$ otherwise. Therefore, by Lemma
\ref{LemmaJoin-preserving}, $h$ preserves arbitrary joins, hence
$g$ is a multi-morphism of $\cCal$.

On the other hand, let $w=f\circ v\in V_T$ and
$\omega\subseteq\l_\beta$. Then
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\[w(a[\vee\omega,\beta])=f(v(a)[\vee
v_\beta(x)\,;\,x\in\omega])=\bigvee_{x\in\omega}w(a[x,\beta])\,
,\]
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
since $f$ is a multi-morphism.\end{proof}

\begin{definition}Let $\l\in\cCal$ and $A\subseteq\l$. We say that $A$
{\it generates} $\l$ if each element of $\l$ is the join of
elements in $A$.\end{definition}

\begin{theorem}
Let $\l\in\cCal(\mathbf{\s})$,
$\{\l_\alpha\in\cCal(\s_\alpha)\}_{\alpha\in\Omega}$, and
$T=\prod_\alpha T_\alpha$ with $T_\alpha\subseteq\aut(\l_\alpha)$.
\begin{enumerate}
%
\item If $\l\in\Sep_T(\l_\alpha,\alpha\in\Omega)$, then there is a
multi-morphism $f:\prod_\alpha\l_\alpha\rightarrow \l$ of $\cCal$
such that $f(\prod_\alpha\l_\alpha)$ generates $\l$, and for all
$g\in\Gamma_\cap$ and $w\in V_T(f)$, there is a unique
$h\in\cCal(\l,2)$ and a unique $u\in\cCal(\l,\l)$ such that the
diagrams
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
$$\bfig
 \ptriangle|alm|/>`>`>/[\prod_\alpha\l_\alpha`\l`2;f`g`!h]
 \efig
 %
 \hspace{1cm}\bfig
 \ptriangle|alm|/>`>`>/[\prod_\alpha\l_\alpha`\l`\l;f`w`!u]
 \efig$$
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
commute.
%
\item Conversely, if there is a multi-morphism of $\cCal$
$f:\prod_\alpha\l_\alpha\rightarrow \l$ satisfying all conditions
of part (1), then there is
$\l_0\in\Sep_T(\l_\alpha,\alpha\in\Omega)$ such that
$\l\cong\l_0$.
\end{enumerate}
\end{theorem}

\begin{proof} (1) By Lemma
\ref{LemmafisaBimorphism}, the map
$f:\prod_\alpha\l_\alpha\rightarrow\l$ defined as
$f(a)=\prod_\alpha a_\alpha$, is a multi-morphism of $\cCal$, and
obviously, $f(\prod_\alpha\l_\alpha)$ generates $\l$. Moreover, by
Axiom P4, for all $w\in V_T(f)$ there is a unique $u$ such that
the second diagram commutes.

Let $g=\cap_\alpha g_\alpha\in\Gamma_\cap$ and
$\mathbf{\s}=\prod_\alpha\s_\alpha$. Define $a_\alpha:=\vee
G_\alpha^{-1}(0)$ and $x:=\cup_\alpha \pr_\alpha(a_\alpha)$ (see
Lemma \ref{LemmaJoin-preserving}). By Axiom P2 $x\in\l$, and by
Lemma \ref{LemmaExitsh_a:L->2} there is $h_x\in\cCal(\l,2)$ such
that for all $p\in\mathbf{\s}$, $h_x(p)=0$ if and only if $p\in
x$. Hence, $h_x\circ f=g$. Let $h\in\cCal(\l,2)$ such that $h\circ
f=g$. Then on atoms $h$ equals $h_x$, therefore $h=h_x$.

(2) Let $\mathbf{\s_0}=\prod_\alpha\s_\alpha$. We denote by $F$
the mapping from $\mathbf{\s_0}$ to $\l$ induced by the
multi-morphism $f$.

(2.1) For $a,\, b\in\prod_\alpha\l_\alpha$, we write $a\leq b$ if
and only if $a_\beta\leq b_\beta$, for all $\beta\in\Omega$. {\bf
Claim}: The following implication holds: $f(a)\subseteq
f(b)\Rightarrow a\leq b$. As a corollary, $f$ is injective. [{\it
Proof}: Suppose that $f(a)\subseteq f(b)$ and that
$a_\beta\not\leq b_\beta$ for some $\beta\in\Omega$. By Lemma
\ref{LemmaExitsh_a:L->2}, there is
$h_{b_\beta}:\l_\beta\rightarrow 2$ such that for all
$p\in\s_\beta$, $h_{b_\beta}(p)=0$ if and only if $p\in b_\beta$.
Let $g_\beta:=h_{b_\beta}$, $g_\alpha:=h_{0_\alpha}$ for all
$\alpha\ne\beta$, and $g:=\cap_\alpha g_\alpha$. By definition,
$g\in \Gamma_\cap$, hence there is $h\in\cCal(\l,2)$ such that
$h\circ f=g$. As a consequence,
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\[1=g_{\beta}(a_\beta)=g(a)=h(f(a))\subseteq
h(f(b))=g(b)=g_\beta(b_\beta)=0\, ,\]
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
a contradiction.]

As a consequence, since $f(\prod_\alpha\l_\alpha)$ generates $\l$,
for all $p\in\mathbf{\s_0}$, $F(p)$ is an atom of $\l$, and the
mapping from $\mathbf{\s_0}$ to $\mathbf{\s}$ induced by $F$
(which we also denote by $F$) is bijective. Moreover, for all
$a\in\prod_\alpha\l_\alpha$,
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\[f(a)=\vee\{F(p)\,;\,p\in \prod_\alpha a_\alpha\}\, .\]
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
Therefore, for all $a,\, b\in\prod_\alpha\l_\alpha$, we have:
$a\leq b\Rightarrow f(a)\subseteq f(b)$ (note that if $\Omega$ is
finite, then this implication follows directly from the fact that
$f$ is a multi-morphism).

(2.2) Let $\l_0\subseteq 2^{\mathbf{\s_0}}$ defined as
$\l_0:=\{F^{-1}(c)\,;\,c\in\l\}$. Then by what precedes,
$\l_0\in\cCal(\mathbf{\s_0})$ and the map
$F^{-1}:\l\rightarrow\l_0$ is bijective and preserves arbitrary
meets, hence also arbitrary joins. It remains to prove that
$\l_0\in\Sep_T(\l_\alpha,\alpha\in\Omega)$, hence to check that
Axioms P2, P3 and P4 hold in $\l_0$. Below, if $g\in\Gamma_\cap$,
then $G$ denotes the map from $\mathbf{\s_0}$ to $2$ induced by
$g$.

(P2) Let $a\in\prod_\alpha\l_\alpha$, $x:=F(\cup_\alpha
\pr_\alpha(a_\alpha))$, and $p\in\mathbf{\s_0}$ such that
$p_\alpha\not\in a_\alpha$ for all $\alpha\in\Omega$. Suppose that
$F(p)\in \vee x$. From Lemma \ref{LemmaExitsh_a:L->2}, for all
$\beta\in\Omega$, there is $h_{a_\beta}\in\cCal(\l_\beta,2)$ such
that for all $p\in\s_\beta$, $h_{a_\beta}(p)=0$ if and only if
$p\in a_\beta$. Let $g=\cap_\alpha h_{a_\alpha}$. By definition,
$g\in \Gamma_\cap$, hence there is $h\in\cCal(\l,2)$ such that
$h\circ f=g$; whence
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\[\begin{split}1=G(p)=h(F(p))\subseteq h(\vee x)&=
\vee \{h(F(p))\,;\,p\in\cup_\alpha\pr_\alpha(a_\alpha)\} \\
%
&=\vee \{G(p)\,;\,p\in\cup_\alpha\pr_\alpha(a_\alpha)\}=0\, ,
%
\end{split}\]
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
a contradiction. As a consequence, $F^{-1}(\vee F(\cup_\alpha
\pr_\alpha(a_\alpha)))=\cup_\alpha \pr_\alpha(a_\alpha)$, hence
$\cup_\alpha \pr_\alpha(a_\alpha)\in\l_0$.

(P3) Let $\beta\in\Omega$, $A\subseteq\s_\beta$, and
$p\in\mathbf{\s_0}$, such that $p[A,\beta]\in\l_0$, {\it i.e.}
there is $c\in\l$ such that $p[A,\beta]=F^{-1}(c)$. Let $q\in \vee
A$ and let $\hat{p}\in\prod_\alpha\l_\alpha$ such that
$\prod_\alpha\hat{p}_\alpha=\{p\}$. Then $p[q,\beta]\in p[\vee
A,\beta]$, hence, since $f$ is multi-morphism, we find that
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\[ F(p[q,\beta])\in F(p[\vee A,\beta])\subseteq\vee F(p[\vee
A,\beta])=f(\hat{p}[\vee A,\beta])=\bigvee_{q\in A}
F(p[q,\beta])\subseteq c\, .\]
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
As a consequence, $q\in A$, therefore $A\in\l_\beta$.

(P4) Let $v\in T$ and $u\in\cCal(\l,\l)$ such that $u\circ
f=f\circ v$. Define $u_0:=F^{-1}\circ u\circ F$. Then
$u_0\in\aut(\l_0)$ and $u_0(p)=v(p)$ for all $p\in\mathbf{\s_0}$.
\end{proof}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Sufficient and necessary conditions for
$\l_1\aertsbig\l_2=\l_1\otfbig\l_2$}\label{SectionSufficientandNecessaryConditionsFor...}

Let $\l_1,\cdots,\l_n\in\cCal$ be DAC-lattices. In this section,
we prove that $\aerts_{i=1}^n\l_i=\otf_{i=1}^n\l_i$ ({\it i.e.}
$\sep(\l_1,\cdots,\l_n)$ has only one element) if and only if
there is $k$ between $1$ and $n$ such that for all $i\ne k$,
$\l_i$ is a power set.

\begin{lemma}\label{LemmaJoininTensorF}
Let $\{\l_\alpha\in\cCal(\s_\alpha)\}_{\alpha\in\Omega}$ and
$\l\in\sep(\l_\alpha,\alpha\in\Omega)$. Let
$\mathbf{\s}=\prod_\alpha\s_\alpha$ and $R\subseteq\mathbf{\s}$.
Define
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\[\vee_\alpha R:=\bigcup_{p\in\mathbf{\s}}p[\vee
R_\alpha[p]]\, .\]
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
Denote by $\vee$ the join in $\l$. Then for any
$f:\mathbb{N}\rightarrow\Omega$ and any $n\in\mathbb{N}$,
$R^n:=\vee_{f(n)}\cdots\vee_{f(1)}R\subseteq\vee R$. Moreover, if
$\vee_\beta R=R$ for all $\beta\in\Omega$, then
$R\in\otf_\alpha\l_\alpha$.
\end{lemma}

\begin{proof} Obviously, for all $\beta\in\Omega$,
$\vee_\beta(\vee_\beta R)=\vee_\beta R$, and by Axioms P2 and P3,
$\vee_\beta \vee R=\vee R$. Therefore, since $R\subseteq\vee R$,
$R^n\subseteq(\vee R)^n=\vee R$.

Moreover, from Definition \ref{DefinitionSeparatedProduct},
$R\in\otf_\alpha\l_\alpha$ if and only if $\vee_\beta R=R$ for all
$\beta\in\Omega$.
\end{proof}

\begin{theorem}
Let $\{\l_\alpha\in\cCal(\s_\alpha)\}_{\alpha\in\Omega}$ and
$\mathbf{\s}=\prod_\alpha\s_\alpha$. If there is at most one
$\beta\in\Omega$ such that $\l_{\beta}\ne 2^{\s_{\beta}}$, then
$\aerts_\alpha\l_\alpha=\otf_\alpha\l_\alpha=\{
R\subseteq\mathbf{\s}\,;\,R_{\beta}[p]\in\l_{\beta},\, \forall
p\in\mathbf{\s}\}$.\end{theorem}

\begin{proof} Let $R\subseteq\mathbf{\s}$.
Since $\aerts_\alpha\l_\alpha\subseteq\otf_\alpha\l_\alpha$,
$\vee_\otfsub R\subseteq\vee_\aertssub R$. We prove that
$\vee_\aertssub R\subseteq\vee_\otfsub R$.

Write $\s_1=\s_{\beta}$, $\l_1=\l_{\beta}$, and
$\s_2=\prod_{\alpha\ne\beta}\s_\alpha$. Hence
$\mathbf{\s}=\s_1\t\s_2$. We denote by
$\pi_i:\mathbf{\s}\rightarrow\s_i$ ($i=1,2$) the $i-$th coordinate
map. From definition \ref{DefinitionSeparatedProduct}, we find
that
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\[\aerts_{\alpha\ne\beta}\l_\alpha=2^{\s_2}\, .\]
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
Therefore, by Proposition \ref{PropositionAssociativity},
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\[\aerts_{\alpha\in\Omega}\l_\alpha=\l_1\aerts 2^{\s_2}\, .\]
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

Let $a\in\l_1$ and $b\in 2^{\s_2}$. Note that $R\subseteq a\t
\s_2\cup \s_1\t b$ if and only if
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\[g(b):=\pi_1(\s_1\t b^c\cap R)\subseteq a\, ,\]
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
where $b^c$ denotes the set-complement of $b$, {\it i.e.}
$b^c=\s_2\backslash b$. As a consequence, we find that
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\[\begin{split}\vee_\aertssub
R&\subseteq\bigcap_{b\subseteq\pi_2(R)}(\vee g(b))\t \s_2\cup \s_1\t b\\
%
&=\bigcup_{f\in 2^{(2^{\pi_2(R)})}}(\cap\{\vee g(b)\,;\,b\in
f^{-1}(1)\})\t(\cap f^{-1}(2))\, .\end{split}\]
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
For $b\subseteq\pi_2(R)$, define
$m(b):=\{a\subseteq\pi_2(R)\,;\,b\nsubseteq a\}$. Then,
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\[\vee_\aertssub R=\bigcup_{b\subseteq\pi_2(R)}(\cap\{\vee
g(a)\,;\,a\in m(b)\})\t b\, .\]
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

Note that for all $q\in b$, $q^c\cap\pi_2(R)\in m(b)$. Moreover,
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\[g(q^c\cap\pi_2(R))=\pi_1(\s_1\t(q\cup \pi_2(R)^c)\cap
R)=R_1[(\cdot,q)]\, .\]
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
As a consequence,
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\[\begin{split}\vee_\aertssub R&\subseteq
\bigcup_{b\subseteq\pi_2(R)}(\cap\{\vee
R_1[(\cdot,q)]\,;\,q\in b\})\t
b\\
%
&=\bigcup_{q\in \pi_2(R)}(\vee R_1[(\cdot,q)])\t q=\vee_{\beta}
R\, .\end{split}\]
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
Finally, by Lemma \ref{LemmaJoininTensorF}, $\vee_{\beta}
R=\vee_\otfsub R$.
\end{proof}

\begin{definition}\label{DefinitionXi}
Let $\s$ be a nonempty set. We denote by $\mo_\s$ the simple
closure space on $\s$ which contains only $\emptyset$, $\s$, and
all singletons of $\s$. We write $\mo_n$ if $\abs\s=n$.

Let $\l\in\cCal$. We say that $\l$ {\it contains $\mo_n$} if there
are $n$ atoms $p_1,\cdots,p_n$ such that $p_1\vee p_n\gtrdot p_i$
for all $i$ between $1$ and $n$.

Let $\{\s_\alpha\}_{\alpha\in\Omega}$ be a family of sets, and
$\mathbf{\s}=\prod_\alpha\s_\alpha$. We denote by
$\Xi(\mathbf{\s})$ the set
$\{R\subseteq\mathbf{\s}\,;\,p_\alpha\ne q_\alpha,\,\forall p,q\in
R,\,\alpha\in\Omega\}$.\end{definition}

\begin{theorem}\label{TheoremPerspectiveThenAertsdiffOtf}
Let $\l_1\in\cCal(\s_1)$ and $\l_2\in\cCal(\s_2)$. If both $\l_1$
and $\l_2$ contain $\mo_3$, then $\l_1\aerts\l_2\ne\l_1\otf\l_2$.
\end{theorem}

\begin{proof}
Let $R=\{p,q,r\}\in\Xi(\s_1\t\s_2)$ such that for $i=1$ and for
$i=2$, $p_i\vee q_i$ covers $p_i$, $q_i$ and $r_i$. By Definition
\ref{DefinitionSeparatedProduct}, $R=\{p,q,r\}\in\l_1\otf\l_2$. On
the other hand, $R\subseteq a\t \s_2\cup \s_1\t b$ if and only if
$p_1\vee q_1\subseteq a$ or $p_2\vee q_2\subseteq b$. As a
consequence,
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\[\vee_\aertssub R=(p_1\vee q_1)\t (p_2\vee q_2)\ne R=\vee_\otfsub
R\, .\]
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\end{proof}

\begin{corollary}
Let $\{\l_i\in\cCal(\s_i)\}_{1\leq i\leq n}$, be DAC-lattices. If
$\aerts_{i=1}^n\l_i=\otf_{i=1}^n\l_i$, then there is at most one
$i$ such that $\l_i\ne 2^{\s_i}$.
\end{corollary}

\begin{proof} Suppose that $\l_i\ne 2^{\s_i}$ for $i=m$ and for $i=k$
with $m\ne k$ between $1$ and $n$, and that
$\aerts_{i=1}^n\l_i=\otf_{i=1}^n\l_i$. Then, for $i=m$ and for
$i=k$, there is an atom $p_i$ which is not a central element.
Therefore, there is an atom $q_i$ such that $p_i\vee q_i$ contains
a third atom (see \cite{Maeda/Maeda:handbook}, Theorems 28.8, 27.6
and Lemma 11.6).

Let $\l_0:=\aerts_{i\ne m}\l_i$ and $\l_1:=\otf_{i\ne m}\l_i$. By
hypothesis, $\l_m\aerts\l_0=\l_m\otf\l_1$. Therefore, from Axioms
P2 and P3, $\l_0=\l_1$. Let $r\in\prod_{i\ne m}\s_i$. From Axiom
P2, $r[p_k]\vee_\aertssub r[q_k]=r[p_k\vee q_k]$, hence contains a
third atom, therefore $\l_0$ contains $\mo_3$. As a consequence,
from Theorem \ref{TheoremPerspectiveThenAertsdiffOtf}, we find
that
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\[\l_m\otf\l_1=\l_m\aerts\l_0\ne\l_m\otf\l_0=\l_m\otf\l_1\, ,\]
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
a contradiction.
\end{proof}

\begin{theorem}\label{TheoremCoatomsCountable}
Let $\l\in\cCal(\s)$ be coatomistic. Suppose that for all
countable set $A$ of coatoms $\l$, $\cup A\ne\s$. For all integer
$i$, let $\l_i=\l$. Then
$\aerts_{i=1}^\infty\l_i\ne\otf_{i=1}^\infty\l_i$.
\end{theorem}

\begin{proof} For all $i$, let $\s_i=\s$.
Denote $\prod_{i=1}^\infty\s_i$ by $\mathbf{\s}$. Let
$R=\{p\in\mathbf{\s}\,;\,p_n=p_m,\,\forall m,n\}\ne\mathbf{\s}$.
From Definition \ref{DefinitionSeparatedProduct},
$R\in\otf_{i=1}^\infty\l_i$. On the other hand, by hypothesis, for
all $a\in\prod_{i=1}^\infty\l_i$, $R\nsubseteq\cup_{i=1}^\infty
\pr_i(a_i)$, hence $\vee_\aertssub R=1$.\end{proof}

\begin{example} The Hypothesis of Theorem
\ref{TheoremCoatomsCountable} is fulfilled for instance if
$\l=\pro(\h)$ with $\h$ a real or complex Hilbert
space.\end{example}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Central elements}\label{SectionCentralElements}

Let $\{\l_\alpha\in\cCal\}_{\alpha\in\Omega}$,
$\l\in\sep(\l_\alpha,\alpha\in\Omega)$ and $\beta\in\Omega$. In
this section we prove that if $z$ is a central element of
$\l_\beta$, then $\pr_\beta(z)$ is a central element of $\l$. As a
corollary, if $\l$ is irreducible, then all $\l_\alpha$'s are
irreducible. We give some sufficient conditions under which the
converse result holds for the bottom element
$\aerts_\alpha\l_\alpha$, and we describe the decomposition into
irreducible components of $\aerts_\alpha\l_\alpha$ in terms of the
decompositions of each $\l_\alpha$.

\begin{definition}
Let $a$ and $b$ be elements of a lattice. Then $(a,b)$ is said to
be a {\it modular pair} (in symbols $(a,b)M$) if $(c\vee a)\land
b=c\vee(a\land b)$ for all $c\leq b$.
\end{definition}

\begin{lemma}\label{LemmaCentralElements}
Let $\l\in\cCal(\s)$ and $z\in\l$. Then $z$ is a central element
of $\l$ if and only if $z^c:=\s\backslash z\in\l$ and $(z,z^c)M$
and $(z^c,z)M$.
\end{lemma}

\begin{proof} Direct from Theorem 4.13 ($\varepsilon$) in
\cite{Maeda/Maeda:handbook}.\end{proof}

\begin{corollary}\label{CorollaryCentralElements}
Let $\l\in\cCal(\s)$ be orthocomplemented and let $z\in\l$. Then
$z$ is a central element of $\l$ if and only if
$z^\p=z^c$.\end{corollary}

\begin{proof}
First, note that $a^\p\subseteq a^c$ for all $a\in\l$. Suppose
that $z$ is a central element of $\l$. Then, since $(z,z^c)M$, we
find that
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\[z^c=(z^\p\vee z)\cap z^c=z^\p\vee(z\cap z^c)=z^\p\, .\]
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

Suppose now that $z^\p=z^c$. We prove that $(z,z^c)M$. The proof
for $(z^c,z)M$ is similar. Let $c\in\l$ with $c\subseteq z^c$.
Then $z=(z^c)^\p\subseteq c^\p$, therefore
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\[(c\vee z)\cap z^c=(c\cup z)^{\p\p}\cap z^\p=[(c^\p\cap z^c)\cup
z]^\p=c^{\p\p}=c\, .\]
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\end{proof}

\begin{theorem} Let
$\{\l_\alpha\in\cCal(\s_\alpha)\}_{\alpha\in\Omega}$,
$\l\in\sep(\l_\alpha,\alpha\in\Omega)$ and $\beta\in\Omega$. If
$z$ is a central element of $\l_\beta$, then $\pr_\beta(z)$ is a
central element of $\l$.\end{theorem}

\begin{proof}
From Lemma \ref{LemmaCentralElements}, $z^c:=\s_\beta\backslash
z\in\l_\beta$. Therefore, by Axiom P2,
$(\pr_\beta(z))^c=\pr_\beta(z^c)\in\l$. We denote $\pr_\beta(z)$
by $Z$ and $\pr_\beta(z^c)$ by $Z^c$. We prove that $(Z,Z^c)M$.
The proof for $(Z^c,Z)M$ is similar.

Let $R\in\l$ with $R\subseteq Z^c$. Let
$\mathbf{\s}=\prod_\alpha\s_\alpha$. Write $R$ as
$R=\cup\{p[R_\beta[p]]\,;\,p\in\mathbf{\s}\}$. By Lemma
\ref{LemmaJoininTensorF}, $R\vee Z=\cup\{p[R_\beta[p]\vee
z]\,;\,p\in\mathbf{\s}\}$. Therefore, since $(z,z^c)$ is modular
pair of $\l_\beta$, we find that
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\[
(R\vee Z)\cap Z^c=\bigcup_{p\in\mathbf{\s}}p[(R_\beta[p]\vee
z)\cap z^c]=\bigcup_{p\in\mathbf{\s}}p[R_\beta[p]]=R\, .\]
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\end{proof}

\begin{corollary} Let $\{\l_\alpha\in\cCal\}_{\alpha\in\Omega}$
and $\l\in\sep(\l_\alpha,\alpha\in\Omega)$. If $\l$ is
irreducible, then all $\l_\alpha$'s are
irreducible.\end{corollary}

\begin{definition}\label{DefinitionCentralElements}
Let $\l\in\cCal(\s)$, $p\in\s$, and $a\in\l$. We denote the center
of $\l$ by $\mathcal{Z}(\l)$, we write
$e(p)=\cap\{z\in\mathcal{Z}(\l)\,;\, p\in z\}$ and
$\mathcal{A}(\mathcal{Z}(\l))=\{e(p)\,;\,p\in\s\}$, and $[0,a]$
denotes the interval $\{b\in\l\,;\, 0\subseteq b\subseteq a\}$.
\end{definition}

\begin{lemma}\label{LemmaDecompositionIrreducibleComponents}
Let $\l\in\cCal(\s)$ be orthocomplemented. Then
$\mathcal{Z}(\l)$ is a complete atomistic sublattice of $\l$, the
set of atoms of which is given by $\mathcal{A}(\mathcal{Z}(\l))$.
Moreover, $\s=\coprod\{z\,;\,z\in\mathcal{A}(\mathcal{Z}(\l))\}$,
and
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\[\l\cong\prod_{z\in\mathcal{A}(\mathcal{Z}(\l))}[0,z]\, ,\]
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
ordered by $f\leq g\Leftrightarrow f(z)\leq g(z)$ for all
$z\in\mathcal{A}(\mathcal{Z}(\l))$.
\end{lemma}

\begin{proof}
Let $\omega\subseteq\mathcal{Z}(\l)$. Then
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\[(\cap\omega)^c\supseteq(\cap\omega)^\p=
\left(\bigcap_{z\in\omega} z^{\p\p}\right)^\p=
\left(\bigcup_{z\in\omega}
z^c\right)^{\p\p}=((\cap\omega)^c)^{\p\p}\, ,\]
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
therefore, $\cap\omega\in\mathcal{Z}(\l)$ by Corollary
\ref{CorollaryCentralElements}. Hence, $\mathcal{Z}$ is a complete
sublattice of $\l$.

Let $p\in\s$ and $r\in e(p)$. Then $e(r)\subseteq e(p)$. Suppose
that $p$ is not in $e(r)$. Then $p\in e(r)^c\cap e(p)$, hence
$e(p)\subseteq e(r)^c\cap e(p)$, therefore $e(p)\subseteq e(r)^c$;
whence $e(r)\subseteq e(p)^c\cap e(p)$, a contradiction. As a
consequence, $p\in e(r)$, therefore $e(p)\subseteq e(r)$, hence
$e(p)=e(r)$.

Let $p,\,q\in\s$. We have proved that either $e(p)=e(q)$, or
$e(p)\cap e(q)=0$. As a consequence,
$\s=\coprod\{z\,;\,z\in\mathcal{A}(\mathcal{Z}(\l))\}$, hence
$\mathcal{A}(\mathcal{Z}(\l))$ is the set of atoms of
$\mathcal{Z}(\l)$, and $\mathcal{Z}(\l)$ is atomistic. The rest of
the proof follows directly from Lemma 5.1 in
\cite{Maeda/Maeda:handbook}.
\end{proof}

\begin{theorem}[\cite{Ischi:2000}, Theorem 1]
Let $\{\l_\alpha\in\cCal(\s_\alpha)\}_{\alpha\in\Omega}$ with for
all $\alpha\in\Omega$, $\l_\alpha$ orthocomplemented. Suppose that
one of the following assumptions holds.
%
\begin{enumerate}
%
\item $\Omega$ is finite.
%
\item For all $\alpha\in\Omega$, $\l_\alpha$ has the covering
property, and for all $p\ne q\in\s_\alpha$ with $e(p)=e(q)$,
$p\vee q$ contains an infinite number of atoms.
%
\end{enumerate}
Then, $\aerts_\alpha\l_\alpha$ is irreducible if and if all
$\l_\alpha$'s are irreducible. Moreover,
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\[\mathcal{A}(\mathcal{Z}(\aerts_\alpha\l_\alpha))=
\{\prod_\alpha Z_\alpha\,;\,
Z\in\prod_\alpha\mathcal{A}(\mathcal{Z}(\l_\alpha))\}\, .\]
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\end{theorem}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Automorphisms}\label{SectionAutomorphisms}

Let $\l_1,\cdots,\l_n\in\cCal$ different from $2$,
$\l\in\sep(\l_1,\cdots,\l_n)$ and $u\in\cCal(\l,\l)$ sending atoms
to atoms. In this section we prove that if $u$ is {\it large} (see
Definition \ref{DefinitionLarge} below), then there is a
permutation $f$ and arrows $v_i\in\cCal(\l_i,\l_{f(i)})$ sending
atoms to atoms such that for any atom $p$ of $\l$,
$u(p)_{f(i)}=v_i(p_i)$. We need some hypotheses on each $\l_i$
which are true for instance if each $\l_i$ is irreducible
orthocomplemented with the covering property or an irreducible
DAC-lattice. Note that our hypotheses imply irreducibility. Note
also that if $u$ is an automorphism, then $u$ is large.

\begin{definition}\label{DefinitionConnected}
Let $\l\in\cCal(\s)$. We say that $\l$ is {\it weakly connected}
if $\l\ne 2$ and if there is a {\it connected covering} of $\s$,
that is a family of subsets
$\{A^\gamma\subseteq\s\,;\,\gamma\in\sigma\}$ such that
%
\begin{enumerate}
\item $\s=\cup \{A^\gamma\,;\,\gamma\in \sigma\}$ and $\abs{
A^\gamma}\geq 2$ for all $\gamma\in\sigma$,
%
\item for all $\gamma\in\sigma$ and all $p\ne q\in A^\gamma$,
$p\vee q$ contains a third atom,
%
\item for all $p,\,q\in\s$, there is a finite subset
$\{\gamma_1,\cdots,\gamma_n\}\subseteq\sigma$ such that $p\in
A^{\gamma_1}$, $q\in A^{\gamma_n}$, and such that
$\abs{A^{\gamma_i}\cap A^{\gamma_{i+1}}} \geq 2$ for all $1\leq
i\leq n-1$.
%
\end{enumerate}
%
We say that $\l$ is {\it connected} if $\l\ne 2$ and for all
$p,\,q\in\s$, $p\vee q$ contains a third atom, say $r$, such that
$p\in q\vee r$ and $q\in p\vee r$.
\end{definition}

\begin{remark}
Note that in part 2 of Definition \ref{DefinitionConnected}, it is
not required that the third atom under $p\vee q$ is in $A^\gamma$.
Let $\l\in\cCal(\s)$. If $\l$ is weakly connected, then $\l$ is
irreducible (follows directly from Lemma
\ref{LemmaCentralElements} and Lemma 5.1 in
\cite{Maeda/Maeda:handbook}, Theorem 4.13). On the other hand, if
$\l\ne 2$ and $\l$ is irreducible orthocomplemented with the
covering property or an irreducible DAC-lattice, then $\l$ is
connected.
\end{remark}

\begin{definition}\label{DefinitionLarge}
Let $\{\l_\alpha\in\cCal(\s_\alpha)\}_{\alpha\in\Omega}$, with for
all $\alpha\in\Omega$, $\l_\alpha\ne 2$. Let
$\l\in\sep(\l_\alpha,\alpha\in\Omega)$,
$\mathbf{\s}=\prod_\alpha\s_\alpha$, and let $u\in\cCal(\l,\l)$.
We say that $u$ is {\it large} if for all $\beta\in\Omega$ and
$p\in\mathbf{\s}$, $u(p[\s_\beta])$ is not an atom and
$u(1)\nsubseteq \pr_\beta(p_\beta)$.\end{definition}

\begin{lemma}\label{LemmaJoinP1diffQ1andP2diffQ2}
Let $\{\l_\alpha\in\cCal(\s_\alpha)\}_{\alpha\in\Omega}$ and
$\l\in\cCal(\mathbf{\s})$ with
$\mathbf{\s}=\prod_\alpha\s_\alpha$. Suppose that Axiom P2 holds
in $\l$. Let $p,\,q\in\mathbf{\s}$.
%
\begin{enumerate}
%
\item If $p_\beta\ne q_\beta$ for at lest two $\beta\in\Omega$,
then $p\vee q=p\cup q$.
%
\item For all $\beta\ne \gamma\in\Omega$ and for all
$b\in\l_\beta$ and $c\in\l_\gamma$ such that $p_\beta\in b$ and
$p_\gamma\in c$, $p[b,\beta]\vee p[c,\gamma]=p[b,\beta]\cup
p[c,\gamma]$.
\end{enumerate}
\end{lemma}

\begin{proof} (1) Define $\Omega_{\ne}:=\{\beta\in\Omega\,;\,
p_\beta\ne q_\beta\}$. Let $r\in p\vee q$, and
$\beta\ne\gamma\in\Omega_{\ne}$. Then
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\[\begin{split}
p,\,q\in[\pr_\beta(p_\beta)\cup
\pr_\gamma(q_\gamma)]&\cap[\pr_\beta(q_\beta)\cup
\pr_\gamma(p_\gamma)]\\
%
&=[\pr_\beta(p_\beta)\cap \pr_\gamma(p_\gamma)]\cup
[\pr_\beta(q_\beta)\cap \pr_\gamma(q_\gamma)]\, .\end{split}\]
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
Therefore, $r_\beta=x_\beta$ and $r_\gamma=x_\gamma$, with $x=p$
or $q$.

(2) We have
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\[\cap\pr_\beta(b)\cap
\pr_\gamma(c)\cap[\pr_\beta(p_\beta)\cup
\pr_\gamma(p_\gamma)]
\bigcap_{\alpha\ne\beta,\gamma}\pr_\alpha(p_\alpha)=p[c,\gamma]\cup
p[b,\beta]\, .\]
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\end{proof}

\begin{theorem}\label{TheoremFactMaps}
Let $\{\l_i\in\cCal(\s_i)\}_{i\in\Omega}$ with $\Omega$ finite
such that for all $i\in\Omega$, $\l_i$ is connected. Let
$\l\in\sep(\l_i,i\in\Omega)$ and $u\in\cCal(\l,\l)$ large, sending
atoms to atoms. Then there is a bijection $f$ of $\Omega$, and for
each $i\in\Omega$, there is $v_i\in \cCal(\l_i,\l_{f(i)})$ sending
atoms to atoms such that $u(p)_{f(i)}=v_i(p_i)$ for all
$p\in\mathbf{\s}$ and $i\in\Omega$.
\end{theorem}

\begin{proof} The proof is similar to the proof of Theorem 3 in
\cite{Ischi:2000}.

(1) Let $p\in\mathbf{\s}$ and $j\in\Omega$. {\bf Claim}: There is
$k \in\Omega$ such that
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\[u(p[\s_j])\subseteq u(p)[\s_k]\, .\]
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
[{\it Proof}\nobreak : Since $\l_j$ is connected, for all
$q_j\in\s_j$ different from $p_j$, $p_j\vee q_j$ contains a third
atom, say $r_j$, and $p_j\in q_j\vee r_j$ and $q_j\in p_j\vee r_j
$. Suppose that $u(p)_k\ne u(p[q_j])_k$ for at lest two indices
$k$. Then, by Lemma \ref{LemmaJoinP1diffQ1andP2diffQ2} part 1 and
Axiom P3,
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\[u(p[r_j])\in u(p\vee p[q_j])=u(p)\vee u(p[q_j])=
u(p)\cup u(p[q_j])\, .\]
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
Assume for instance that $u(p[r_j])=u(p)$. Then,
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\[u(p[q_j])\in u(p\vee p[r_j])=u(p)\vee u(p[r_j])=u(p)\, ,\]
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
a contradiction.]

(2) Let $j\in\Omega$. Denote the $k$ of part 1 by $f(j,p)$. {\bf
Claim}: The map $p\mapsto f(j,p)$ is constant. [{\it
Proof}\nobreak : Let $p,\,q\in\mathbf{\s}$ that differ only by one
component, say $j'\ne j$, {\it i.e.} $p_{j'}\ne q_{j'}$, and
$p_i=q_i$, for all $i\ne j'$. Suppose that $f(j,p)\ne f(j,q)$.
Write $k:=f(j,p)$ and $k':=f(j,q)$.

(2.1) We first prove that
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\[u(p[\s_j ])\vee u(q[\s_j])=u(p[\s_j])\cup u(q[\s_j])\, .\]
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
By hypothesis,
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\[u(p[\s_j])\subseteq u(p)[\s_k]\ \mbox{and}\ u(q[\s_j])\subseteq
u(q)[\s_{k'}]\, .\]
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
Hence, for all $r_j\in\s_j$, we have
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\[u(p)_l=u(p[r_j])_l\,\forall l\ne k\ \mbox{and}\ u(q)_m=
u(q[r_j])_m\,\forall
m\ne k'\, .\]
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

Since $u$ is large, $u(p[\s_j])$ is not an atom, hence there is
$r_j\in\s_j$ such that
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\[u(p[r_j])_k\ne u(q)_k\, .\]
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
Note that $u(q)_k=u(q[r_j])_k$. Therefore, since $p[r_j]$ and
$q[r_j]$ differ only by one component, by part 1 we have
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\[u(p[r_j])_l=u(q[r_j])_l\,\forall l\ne k\, .\]
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
As a consequence,
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\[u(q)_l=u(q[r_j])_l=u(p[r_j])_l=u(p)_l\,\forall l\ne k,\,k'\, ,\]
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
and
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\[u(p)_{k'}=u(p[r_j])_{k'}=u(q[r_j])_{k'}\in\pi_{k'}(u(q[\s_j]))\,
.\]
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

On the other hand, since $u(q[\s_j])$ is not an atom, there is
$s_j\in\s_j$ such that
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\[u(q[s_j])_{k'}\ne u(p)_{k'}\, .\]
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
Note that $u(p)_{k'}=u(p[r_j])_{k'}$. Therefore, since $q[s_j]$
and $p[s_j]$ differ only by on component, by part 1 we have
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\[u(q[s_j])_l=u(p[s_j])_l\,\forall l\ne k'\, .\]
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
As a consequence,
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\[u(q)_k=u(q[s_j])_k=u(p[s_j])_k\in\pi_k(u(p[\s_j]))\, .\]
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

To summarize, we have proved that $u(p)_l=u(q)_l$ for all $l\ne
k,\,k'$, $u(p)_{k'}\in\pi_{k'}(u(q[\s_j]))$, and $u(q)_k\in
\pi_k(u(p[\s_j]))$. As a consequence, the statement follows from
Lemma \ref{LemmaJoinP1diffQ1andP2diffQ2} part 2.

(2.2) Since $\l_{j'}$ is connected, there is $s_{j'}\in p_{j'}\vee
q_{j'}$ such that $p_{j'}\in q_{j'}\vee s_{j'}$ and $q_{j'}\in
p_{j'}\vee s_{j'}$. Let $r=p[s_{j'}]$. Then, by Axiom P3, for all
$t_j\in\s_j$, $r[t_j]\subseteq p[t_j]\vee q[t_j]$. Therefore
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\[r[\s_j]\subseteq p[\s_j]\vee q[\s_j]\, ,\]
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
and for the same reason
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\[p[\s_j]\subseteq q[\s_j]\vee r[\s_j]\ \mbox{and}\ q[\s_j]\subseteq
p[\s_j]\vee r[\s_j]\, .\]
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
Now, by part 2.1,
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\[u(p[\s_j]\vee q[\s_j])=u(p[\s_j])\vee u(q[\s_j])=u(p[\s_j])
\cup u(q[\s_j])\, .\]
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
As a consequence, by part 1, $u(r[\s_j])\subseteq u(p[\s_j])$ or
$u(r[\s_j])\subseteq u(q[\s_j])$. Assume for instance that
$u(r[\s_j])\subseteq u(p[\s_j])$. Then,
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\[u(q[\s_j])\subseteq  u(p[\s_j ]\vee r[\s_j])=u(p[\s_j])\vee
u(r[\s_j])=u(p[\s_j])\, ,\]
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
a contradiction. Hence, we have proved that if
$p,\,q\in\mathbf{\s}$ differ only by one component, then
$f(j,p)=f(j,q)$.

(2.3) Suppose now that $p$ and $q$ differ by more than one
component. Since $\Omega$ is finite, there is
$s_1,\cdots,s_n\in\mathbf{\s}$ such that $s^1=q$, $s^n=p$, and
such that for all $1\leq i\leq n-1$, $s^i$ and $s^{i+1}$ differ
only by one component. Therefore,
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\[f(j,q)=f(j,s^1)=f(j,s^2)=\cdots=f(j,s^n)=f(j,p)\, ,\]
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
and we are done.]

(3) Let $p_0\in\mathbf{\s}$. Define $f:\Omega\rightarrow\Omega$ as
$f(i):=f(i,p_0)$. Note that by part 2, $f$ does not depend on the
choice of $p_0$. {\bf Claim}: The map $f$ is surjective. [{\it
Proof}\nobreak : Let $k\in\Omega$. Suppose that for all
$i\in\Omega$ $f(i)\ne k$. Let $p,\, q\in\mathbf{\s}$ that differ
only by one component, say $j$. Then $p[\s_j]=q[\s_j]$, therefore
$u(p[\s_j])=u(q[\s_j])$. Moreover, by part 1, there is $k'\ne k$
such $u(p[\s_j ])\subseteq u(p)[\s_{k'}]$. As a consequence,
$u(p)_k=u(q)_k$.

Since $\Omega$ is finite, by the same argument as in part 2.3, we
find that $u(p)_k=u(q)_k$, for all $p,\,q\in\mathbf{\s}$. As a
consequence, $u(\mathbf{\s})\subseteq \pr_k(u(p_0)_k)$, a
contradiction since $u$ is large.]

(4) Let $p_0\in\mathbf{\s}$ and $j\in\Omega$. Define $v_j :\l_j
\rightarrow\l_{f(j )}$ as $v_j (a_j ):=\pi_{f(j )}(u(p_0[a_j ]))$.
{\bf Claim}: $v_j $ does not depend on the choice of $p_0$. [{\it
Proof}\nobreak : Let $q\in\mathbf{\s}$ that differs from $p_0$
only by one component, say $j'\ne j $. Then, by Axiom P3, we have
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\[\begin{split}\pi_{f(j )}(u(q[a_j ]))&=
\pi_{f(j)}(u(\vee \{q[r_j]\,;\,r_j\in
a_j\}))\\
%
&=\pi_{f(j)}(\vee \{u(q[r_j])\,;\,r_j\in a_j\})=\bigvee_{r_j\in
a_j} u(q[r_j])_{f(j)}\, ,\end{split}\]
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
and the same formula holds for $\pi_{f(j)}(u(p_0[a_j]))$. Now
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\[u(q[r_j ])_{f(j )}=\pi_{f(j )}(u(s[\s_{j'}]))=u(p_0[r_j ])_{f(j
)}\, ,\]
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
where $s=q[r_j]$. As a consequence,
$\pi_{f(j)}(u(q[a_j]))=\pi_{f(j)}(u(p_0[a_j]))$.

Since $\Omega$ is finite, by the same argument as in part 2.3, we
find that $\pi_{f(j )}(u(q[a_j ]))$ $=\pi_{f(j )}(u(p_0[a_j ]))$
for all $q\in\mathbf{\s}$.]

It remains to check that $v_j$ preserves arbitrary joins. Let
$\omega\subseteq\l_j$. Then, by Axiom P3,
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\[\begin{split}
v_j(\vee\omega)&=\pi_{f(j)}(u(p[\vee\omega]))=
\pi_{f(j)}(u(\vee\{p[x,j]\,;\,x
\in\omega\}))\\
%
&=\pi_{f(j)}(\vee\{u(p[x,j])\,;\,x\in\omega\})\\
%
&=\bigvee_{x\in\omega}\pi_{f(j)}(u(p[x,j]))=
\bigvee_{x\in\omega}v_j(x)\,
.\end{split}\]
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\end{proof}

\begin{corollary}
If the $u$ in Theorem \ref{TheoremFactMaps} is an automorphism,
then all $v_i$'s are isomorphisms.\end{corollary}

\begin{theorem}\label{TheoremAutoFactorwithWeaklyConnected} If the
$u$ in Theorem \ref{TheoremFactMaps} is an automorphism, the
statement remains true if for all $i\in\Omega$, $\l_i$ is weakly
connected.\end{theorem}

\begin{proof}
The proof is similar as in Theorem \ref{TheoremFactMaps}. We only
sketch the arguments that must be modified. Note that since $u$ is
an automorphism and $\l_i\ne 2$ for all $i\in\Omega$, $u$ is
large.

(1) Since $\l_j$ is weakly connected, there is
$\gamma_0\in\sigma^j$ such that $p_j\in A_j^{\gamma_0}$. By
hypothesis and Axiom P3, for all $q_j\in A_j^{\gamma_0}$, $p\vee
(p[q_j])$ contains a third atom, hence also $u(p)\vee u(p[q_j])$
since $u$ is injective. As a consequence, there is
$k_{\gamma_0}\in\Omega$ such that $u(p[A_j^{\gamma_0} ])\subseteq
u(p)[\s_{k_{\gamma_0}}]$. Moreover, since $u$ is injective, by the
third hypothesis in Definition \ref{DefinitionConnected}, the map
$\gamma\mapsto k_\gamma$ is constant. Therefore, since $\cup
A^\gamma_j=\s_j$, we find that $u(p[\s_j ])\subseteq u(p)[\s_k ]$.

(2) Take $p,\,q\in\mathbf{\s}$ that differ only by one component
such that $q_{j'}$ and $p_{j'}$ are in the same $A_j^\gamma$.

(2.2) By hypothesis, $p_{j'}\vee q_{j'}$ contains a third atom,
say $r_{j'}$, therefore $r[\s_j]\subseteq p[\s_j]\vee q[\s_j]$,
hence, $u(r[\s_j])=u(p[\s_j])$ or $u(r[\s_j])=u(q[\s_j])$, a
contradiction since $u$ is injective. As a consequence,
$f(j,q)=f(j,p)$. Now, since $u$ is injective, by the third
hypothesis in definition \ref{DefinitionConnected}, we find that
$f(j,q)=f(j,p)$, for all $p,\,q\in\mathbf{\s}$ that differ only by
the component $j'$.
\end{proof}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Orthocomplementation}\label{SectionOrthocomplementation}

This section is devoted to our main result, namely that if
$\l\in\sep(\l_1,\cdots,\l_n)$ is orthocomplemented, then
$\l=\aerts_{i=1}^n\l_i$. We provide two proofs. The first one uses
Theorem \ref{TheoremAutoFactorwithWeaklyConnected}, hence requires
each $\l_i$ to be weakly connected, so in particular irreducible.
In addition we must assume that each $\l_i$ is coatomistic and
that $\l$ is transitive. For the second proof, we assume that each
$\l_i$ is orthocomplemented and that all its irreducible
components different from $2$ are weakly connected, but we do not
need to assume that $\l$ is transitive.

\begin{definition} Let $\{\s_\alpha\}_{\alpha\in\Omega}$ be a
family of nonempty sets, $p\in\mathbf{\s}=\prod_\alpha\s_\alpha$,
$\beta,\,\gamma\in\Omega$, $A_\beta\subseteq\s_\beta$, and
$A_\gamma\subseteq\s_\gamma$. Then
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\[ p[A_\beta,A_\gamma]:=\{q\in\mathbf{\s}\,;\,
q_\beta\in A_\beta,\,q_\gamma\in A_\gamma,\,
q_\alpha=p_\alpha,\,\forall \alpha\ne \beta,\, \gamma\}\, .\]
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\end{definition}

\begin{lemma}\label{LemmaCoatomsAertsCoatomsSep}
Let $\{\l_i\in\cCal(\s_i)\}_{1\leq i\leq n}$,
$x\in\prod_{i=1}^n\l_i$ with $x_i$ coatoms, and let
$\l\in\sep(\l_i,1\leq i\leq n)$. Then $X:=\cup_{i=1}^n \pr_i(x_i)$
is a coatom of $\l$.
\end{lemma}

\begin{proof}
By Axiom P2, $X\in\l$. Let $\mathbf{\s}=\prod_{i=1}^n\s_i$ and
$p\in\mathbf{\s}$ not in $X$. Write $R^0:=p\cup X$ and $y:=p\vee
X$. Define $R^N:=\vee_n\cdots\vee_1 R^{N-1}$. By Lemma
\ref{LemmaJoininTensorF}, $R^N\subseteq y$, for all $N$. Now,
$\vee_k R^0= p[\s_k]\cup R^0$, thus $R^1=R^0\cup_{k=1}^n p[\s_k]$.
Further,
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\[R^2=R^0\bigcup_{k_1=1}^{n-1}
\bigcup_{k_2=k_1+1}^n p[\s_{k_1},\s_{k_2}]\, .\]
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
Hence obviously, for $N=n$, $R^N=\mathbf{\s}$.
\end{proof}

\begin{proposition} Let
$\{\l_\alpha\in\cCal(\s_\alpha)\}_{\alpha\in\Omega}$.
%
\begin{enumerate}
%
\item If $\aerts_\alpha\l_\alpha$ is coatomistic, then all
$\l_\alpha$'s are coatomistic.
%
\item Suppose that $\Omega$ is finite and that all $\l_\alpha$'s
are weakly connected. If $\aerts_\alpha\l_\alpha$ is
orthocomplemented, then all $\l_\alpha$'s are orthocomplemented.
%
\end{enumerate}
%
\end{proposition}

\begin{proof}
(1) Let $\mathbf{\s}=\prod_\alpha\s_\alpha$ and $p\in\mathbf{\s}$.
From Definition \ref{DefinitionSeparatedProduct}, a coatom of
$\aerts_\alpha\l_\alpha$ over $p$ is necessarily of the form
$\cup_\alpha\pr_\alpha(x_\alpha)$ where all $x_\alpha$'s are
coatoms and with $p_\beta\in x_\beta$ for some $\beta\in\Omega$.
Let $\s'[p_\alpha]$ denote the set of coatoms of $\l_\alpha$ above
$p_\alpha$ (write $\s_\alpha':= \s_\alpha'[0_\alpha]$). If
$\aerts_\alpha\l_\alpha$ is coatomistic, then
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\[p=\bigcap\{\cup_\alpha\pr_\alpha(x_\alpha)\,;\,
x\in\prod_\alpha\s'_\alpha\ \mbox{with}\ p_\beta\in x_\beta\
\mbox{for\ some}\ \beta\in\Omega\}\, .\]
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
As a consequence, all $\l_\alpha$'s are coatomic. Moreover, for
all $\alpha\in\Omega$,
$p_\alpha=\cap\{x_\alpha\,;\,x_\alpha\in\s'[p_\alpha]\}$, hence
$\l_\alpha$ is coatomistic.

(2) Let $\beta\in\Omega$ and $a\in\l_\beta$. We first prove that
there is $b\in\l_\beta$ such that $\pr_\beta(a)'=\pr_\beta(b)$.
Let $f:\mathbf{\s}\rightarrow\prod\s_\alpha'$ such that
$p'=\cup_\alpha\pr_\alpha(f_\alpha(p))$ for all $p\in\mathbf{\s}$.

(2.1) Let $p,\, q\in\mathbf{\s}$. {\bf Claim}: If $f(p)_\alpha\ne
f(q)_\alpha$ for at least two $\alpha\in\Omega$, then $p\vee
q=p\cup q$. [{\it Proof}\nobreak : Let
$\Omega_{\ne}=\{\alpha\in\Omega\,;\,f(p)_\alpha\ne f(q)_\alpha\}$
and $\gamma\ne\delta\in\Omega_{\ne}$. Let $r\in p\vee q$. Then,
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\[\cup_\alpha\pr_\alpha(f(r)_\alpha)=r'\supseteq p'\cap
q'\supseteq \pr_\gamma(f(p)_\gamma)\cap\pr_\delta(f(q)_\delta)\cup
\pr_\gamma(f(q)_\gamma)\cap\pr_\delta(f(p)_\delta) \, .\]
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
Therefore, $f(r)_\gamma=x_\gamma$ and $f(r)_\delta=x_\delta$ with
$x=f(p)$ or $x=f(q)$. As a consequence, $r'=p'$ or $r'=q'$, hence
$r=p$ or $r=q$.]

(2.2) {\bf Claim}: Let $p,\, q\in\mathbf{\s}$ such that $p$ and
$q$ differ only by one component, say $\gamma$. Then
$f(p)_\alpha=f(q)_\alpha$ for all $\alpha\ne\gamma$. [{\it
Proof}\nobreak : Since $\l_\gamma$ is weakly connected, from part
2.1 we find that there is $\delta\in\Omega$ such that for all
$r\in\s_\gamma$, $f(p[r,\gamma])_\alpha=f(p)_\alpha$ for all
$\alpha\ne\delta$. Hence,
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\[p[\s_\gamma]'=\bigcup_{\alpha\ne
\delta}\pr_\alpha(f(p)_\alpha)\cup\pr_\delta(x)\, ,\]
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
for some $x\in\l_\delta$. Now $p[\s_\gamma]\cap
\pr_\gamma(f(p)_\gamma)\ne 0$. Therefore, if $\delta\ne\gamma$,
then $p[\s_\gamma]\cap p[\s_\gamma]'\ne 0$, a contradiction. As a
consequence, $\delta=\gamma$ and $x=0$.]

Let $p,\,q\in\mathbf{\s}$ such that $p\in\pr_\beta(a)'$ and such
that $p$ and $q$ differ only by one component, say $\gamma$,
different from $\beta$. Then $\pr_\beta(a)\subseteq
p'=\cup_\alpha\pr_\alpha(f(p)_\alpha)$, hence $a\subseteq
f(p)_\beta$. Now, from part 2.2, $f(q)_\beta=f(p)_\beta$,
therefore $\pr_\beta(a)\subseteq q'$, thus $q\in\pr_\beta(a)'$.
Now, by the same argument as in the proof of Theorem
\ref{TheoremFactMaps} part 2.3, we find that for all
$q\in\mathbf{\s}$ such that $q_\beta=p_\beta$,
$q\in\pr_\beta(a)'$.

As a consequence, we have proved that there is an element in
$\l_\beta$, which we denote by $a^{\p_\beta}$, such that
$\pr_\beta(a)'=\pr_\beta(a^{\p_\beta})$. Obviously, the mapping
$\p_\beta:\l_\beta\rightarrow\l_\beta$ is an orthocomplementation.
\end{proof}

\begin{theorem}
Let $\{\l_i\in\cCal(\s_i)\}_{1\leq i\leq n}$, with for all $i$,
$\l_i$ coatomistic and weakly connected, and let
$\l\in\sep(\l_i,1\leq i\leq n)$. If $\l$ is transitive and
orthocomplemented, then $\l=\aerts_{i=1}^n\l_i$.
\end{theorem}

\begin{proof} Let $\mathbf{\s}=\prod_{i=1}^n\s_i$. We denote the
orthocomplementation of $\l$ by $'$. Let $x\in\prod_{i=1}^n\l_i$
with $x_i$ coatoms, and $X:=\cup_{i=1}^n \pr_i(x_i)$. By Lemma
\ref{LemmaCoatomsAertsCoatomsSep}, $X'=p$, for some
$p\in\mathbf{\s}$. Let $q\in\mathbf{\s}$. Since $\l$ is
transitive, there is $u\in\aut(\l)$ such that $u(p)=q$, hence
$q'=u(p)'$. Define $u'\in\aut(\l)$ as $u'(a):=u(a')'$. Then
$q'=u(p)'=u'(p')=u'(X)$. From Theorem
\ref{TheoremAutoFactorwithWeaklyConnected}, $u'$ factors,
therefore there is $y\in\prod_{i=1}^n\l_i$ with $y_i$ coatoms such
that $q'=Y:=\cup_{i=1}^n \pr_i(y_i)$.
\end{proof}

\begin{remark}
A similar result, using a different set of axioms than Axioms
P1-P3, is established in \cite{Ischi(rmp):2004} for $n=2$ and
$\l_i=\pro(\h_i)$ with $\h_i$ complex Hilbert spaces. If
$\l\in\Sep_T(\l_i,1\leq i\leq n)$, $\l_i$ are transitive, and the
$u$ in Axiom P4 is an ortho-isomorphism of $\l$ for all $v_i\in
T_i$, then the proof does not require Theorem
\ref{TheoremAutoFactorwithWeaklyConnected}. Below we give a second
proof which requires neither Theorem
\ref{TheoremAutoFactorwithWeaklyConnected} nor $\l$ to be
transitive.
\end{remark}

\begin{theorem}
Let $\{\l_i\in\cCal(\s_i)\}_{1\leq i\leq n}$, with for all $i$,
$\l_i$ orthocomplemented. Suppose moreover that for all $i$, all
irreducible components of $\l_i$ different from $2$ are weakly
connected. Let $\l\in\sep(\l_i,1\leq i\leq n)$. If $\l$ is
orthocomplemented, then $\l=\aerts_{i=1}^n\l_i$.
\end{theorem}

\begin{proof} Let $\mathbf{\s}=\prod_{i=1}^n\s_i$. We denote
the orthocomplementation of $\l$ by $'$. From Lemma
\ref{LemmaCoatomsAertsCoatomsSep}, we can define a map $\phi$ on
$\mathbf{\s}$ as $\phi(p)=p^{\#'}$ (see Definition
\ref{DefinitionOrthAerts}). Note that $\phi$ is injective. We
prove in four steps that $\phi$ is surjective.

(1) Let $p\in\mathbf{\s}$ and $a_j\in\l_j$ for some $j$ between
$1$ and $n$. {\bf Claim}: $\phi(p[a_j])=p[a_j]^{\#'}$. [{\it
Proof}\nobreak : If $q\in p[a_j]$, then $p[a_j]^\#\subseteq q^\#$,
hence $\phi(q)\in p[a_j]^{\#'}$. On the other hand, if $q\in
p[a_j]^{\#'}$, then,
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\[p[a_j]^\#=\bigcup_{i\ne j} \pr_i(p_i^{\p_i})\cup
\pr_j(a_j^{\p_j})\subseteq
q'\, .\]
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

Let $r\in q'$ not in $p[a_j]^\#$. Note that
$r[a_j^{\p_j}]\subseteq p[a_j]^\#$. Therefore, by Lemma
\ref{LemmaJoininTensorF}, $r[a_j^{\p_j}\vee r_j]\subseteq q'$. Let
$k\ne j$ between $1$ and $n$. Then, $r[p_k^{\p_k},a_j^{\p_j}\vee
r_j]\subseteq p[a_j]^\#$. Hence, since $r[a_j^{\p_j}\vee
r_j]\subseteq q'$, by Lemma \ref{LemmaJoininTensorF},
$r[\s_k,a_j^{\p_j}\vee r_j]\subseteq q'$. Repeating this argument
as in the proof of Lemma \ref{LemmaCoatomsAertsCoatomsSep}, we
find that $\pr_j(a_j^{\p_j}\vee r_j)\subseteq q'$. As a
consequence, there is an atom $s_j$ with $a_j^{\p_j}\subseteq
s_j^{\p_j}$ such that
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\[q'=\bigcup_{i\ne j}\pr_i(p_i^{\p_i})\cup \pr_j(s_j^{\p_j})\, .\]
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
Therefore, $q=p[s_j]^{\#'}$ with $s_j\in a_j$, hence
$q\in\phi(p[a_j])$.]

(2) Let $p\in\mathbf{\s}$. {\bf Claim}: For all $j$ between $1$
and $n$ there is $q\in \prod_{i=1}^n e(p_i)$ and $k$ between $1$
and $n$ such that $\phi(p[e(p_j)])\subseteq q[e(p_k)]$ (see
Definition \ref{DefinitionCentralElements}). [{\it Proof}\nobreak
: Note that if $a\in\l$, then $a'\subseteq a^c$, where $a^c$
denotes the set-complement of $a$, {\it i.e.}
$a^c=\mathbf{\s}\backslash a$. By part 1,
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\[\phi(p[e(p_j)])\subseteq (p[e(p_j)]^\#)^c=e(p_j)
\mbox{$\prod$}_{i\ne j}
((p_i^{\p_i})^c)\subseteq\mbox{$\prod$}_{i=1}^n e(p_i)\, ,\]
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
since if $q_i\in e(p_i)$, then $(q_i^{\p_i})^c\subseteq e(p_i)$.
If $[0,e(p_j)]=2$ ({\it i.e.} $e(p_j)=p_j$), then the proof is
finished.

Otherwise, let $t_j\ne s_j\in e(p_j)$. Then,
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\[\begin{split}q\in\phi(p[t_j])\vee\phi(p[s_j])&\Leftrightarrow
p[t_j]^\#\cap p[s_j]^\#\subseteq q'\\
%
&\Leftrightarrow \bigcup_{i\ne j}\pr_i(p_i^{\p_i})\cup
\pr_j((t_j\vee s_j)^{\p_j}) \subseteq q'\, ,\end{split}\]
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
hence if and only if $q=\phi(p[r_j])$ for some $r_j\in t_j\vee
s_j$. Therefore, if $t_j\vee s_j$ contains a third atom, so does
$\phi(p[t_j])\vee \phi(p[s_j])$. Hence, by Lemma
\ref{LemmaJoinP1diffQ1andP2diffQ2} part 1, $\phi(p[t_j])$ and
$\phi(p[s_j])$ differ only by one component, say $k$. As a
consequence, since $\l_j$ is weakly connected, for all
$A_j^\gamma$ in the connected covering of $\s_j$, there is
$q^\gamma\in \prod_{i=1}^n e(p_i)$ and $k_\gamma$ such that
$\phi(p[e(p_j)\cap A_j^\gamma])\subseteq
q^\gamma[e(p_{k_\gamma})]$. From Hypotheses 1 and 3 in Definition
\ref{DefinitionConnected}, the maps $\gamma\mapsto k_\gamma$ and
$\gamma\mapsto q_\gamma$ are constant since $\phi$ is injective.]

(3) {\bf Claim}: For all $p\in\mathbf{\s}$ and all $j$ between $1$
and $n$, there is $q\in\prod_{i=1}^n e(p_i)$ such that
$\phi(p[e(p_j)])=q[e(p_j)]$. [{\it Proof}\nobreak : From part 2,
there is $q\in\prod_{i=1}^n e(p_i)$, $k$ between $1$ and $n$, and
$b_k\subseteq e(p_k)$, such that $\phi(p[e(p_j)])= q[b_k]$.

Assume first that $k\ne j$. Let $R^0:=q[b_k]\cup p[e(p_j)]^\#$. By
part 1,
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\[{R^0}'=q[b_k]'\cap p[e(p_j)]^{\#'}=\phi(p[e(p_j)])'\cap
\phi(p[e(p_j)])=0\, ,\]
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
hence $\vee R^0=1$. On the other hand,
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\[
R^0=q[b_k]\bigcup_{i\ne j} \pr_i(p_i^{\p_i})\cup
\pr_j(e(p_j)^c)\subseteq \pr_j(r_j^{\p_j})\bigcup_{i\ne
j}\pr_i(p_i^{\p_i})\, ,
\]
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
for all $r_j\in e(p_j)\cap q_j^{\p_j}$, whence by Axiom P2, $\vee
R^0\ne 1$, a contradiction. As a consequence, $k=j$.

Let $R:=q[b_j]\cup p[e(p_j)]^\#$. From part 1,
$R'=\phi(p[e(p_j)])'\cap \phi(p[e(p_j)])=0$. Therefore, $\vee
R=1$. Now,
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\[R\subseteq\bigcup_{i\ne j}\pr_i(p_i^{\p_i})\cup
\pr_j(e(p_j)^c\vee b_j)\,
.\]
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
Note that for all $a\in\l_i$ with $a\subseteq e(p_j)$, we have
$e(p_j)\cap(a\vee e(p_j)^{\p_j})=a$ (see Corollary
\ref{CorollaryCentralElements}). Therefore, $a\vee
e(p_j)^{\p_j}=a\cup e(p_j)^{\p_j}$. Hence, we find that
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\[R\subseteq\bigcup_{i\ne j}\pr_i(p_i^{\p_i})\cup
\pr_j(e(p_j)^c\cup b_j)\, .\]
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
As a consequence, since $\vee R=1$, from Axiom P2 we find that
$b_j=e(p_j)$.]

(4) Let $p\in\mathbf{\s}$ and $s\in \prod_{i=1}^n e(p_i)$. By part
3, $\phi(p[e(p_1)])=q^1[e(p_1)]$. Therefore, there is $r_1\in
e(p_1)$ such that $\phi(p[r_1])_1=s_1$. Let $k\leq n$ and $r_1\in
e(p_1), \cdots,\,r_k\in e(p_k)$ such that
$\phi(p[r_1,\cdots,r_k])_i=s_i$, for all $i\leq k$, and such that
$\phi(p[r_1,\cdots,r_k])_{k+1}$ is different from $s_{k+1}$. By
part 3,
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\[\phi(p[r_1,\cdots,r_k,e(p_{k+1})])=q^{k+1}[s_1,\cdots
s_k,e(p_{k+1})]\, .\]
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
Hence there is $r_{k+1}\in e(p_{k+1})$ such that
$\phi(p[r_1,\cdots,r_{k+1}])_i=s_i$, for all $i$ between $1$ and
$k+1$. As a consequence, from Lemma
\ref{LemmaDecompositionIrreducibleComponents}, $\phi$ is
surjective.
\end{proof}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Covering property}\label{SectionCoevringProperty}

In this Section we prove that the top element
$\otf_\alpha\l_\alpha$ has the covering property if and only at
most one $\l_\alpha$ is not a power set. We reproduces the
analogue result concerning the bottom element
$\aerts_\alpha\l_\alpha$ which is due to Aerts \cite{Aerts:1982}.
Moreover, for $\l_i=\mo_{\s_i}$ ($i=1,2$) and
$T=\aut(\l_1)\t\aut(\l_2)$, we prove that there is only one
$\l\in\Sep_T(\l_1,\l_2)$ with the covering property.

\begin{theorem}[D. Aerts, \cite{Aerts:1982}]
Let $\{\l_\alpha\in\cCal(\s_\alpha)\}_{\alpha\in\Omega}$ with for
all $\alpha\in\Omega$, $\l_\alpha$ orthocomplemented. If
$\aerts_\alpha\l_\alpha$ has the covering property or is
orthomodular, then there is at most one $\beta\in\Omega$ such that
$\l_{\beta}\ne 2^{\s_{\beta}}$.
\end{theorem}

\begin{proof} Let $\l\in\cCal(\s)$ be orthocomplemented and let
$p,\, q\in\s$
such that $p\vee q=p\cup q$. Define $x:=q^\p\cap(p\vee q)$, then
$0\subseteq x\subseteq p$. If $\l$ has the covering property, then
$1\ne x^\p$, whereas if $\l$ is orthomodular, $x\vee q=p\vee q$.
As a consequence, $x=p$, hence $p\p q$.

Let $\beta\in\Omega$. Suppose that $\l_{\beta}\ne 2^{\s_{\beta}}$.
Then there are two non orthogonal atoms, say $r_{\beta}$ and
$s_{\beta}$. Let $r_\gamma,\,s_\gamma\in\s_\gamma$ for some
$\gamma\in\Omega$ different from $\beta$, and let $p,\,
q\in\prod_\alpha\s_\alpha$ defined as $p_\alpha=q_\alpha$, for all
$\alpha\ne \beta,\,\gamma$, and $p_\alpha=r_\alpha$ and $q_\alpha=
s_\alpha$ if $\alpha=\beta$ or $\gamma$. By Lemma
\ref{LemmaJoinP1diffQ1andP2diffQ2}, $p\vee q=p\cup q$. Therefore,
by what precedes, since $\#$ is an orthocomplementation of
$\aerts_\alpha\l_\alpha$, $p\# q$, hence by Definition
\ref{DefinitionOrthAerts}, $r_\gamma\p_\gamma s_\gamma$. As a
consequence, $\l_\gamma=2^{\s_\gamma}$.\end{proof}

\begin{proposition} Let $\{\l_\alpha\in
\cCal(\s_\alpha)\}_{\alpha\in\Omega}$.
If $\otf_\alpha\l_\alpha$ has the covering property, then all
$\l_\alpha$'s have the covering property.\end{proposition}

\begin{proof} Let $a_\beta\in\l_\beta$, $q_\beta\in\s_\beta$
not in $a_\beta$,
and $p\in\prod_\alpha\s_\alpha$. By the covering property, we find
that $p[q_\beta]\vee p[a_\beta]\gtrdot p[a_\beta]$; Whence by
Axiom P3, $q_\beta\vee a_\beta\gtrdot a_\beta$.\end{proof}

\begin{remark}In the next theorem, we assume that each
$\l_\alpha\in\cCal(\s_\alpha)$ with $\l_\alpha\ne 2^{\s_\alpha}$
contains $\mo_4$. This is for instance the case if $\l_\alpha$ is
orthocomplemented orthomodular with the covering property. Indeed,
if $\l_\alpha\ne 2^{\s_\alpha}$, there is an atom $p_\alpha$ which
is not central, hence such that $e(p_\alpha)$ contains at least
two atoms, say $r$ and $s$. Moreover, $r\vee s$ contains at least
three atoms, for $\l_\alpha$ has the covering property. Finally,
since $\l_\alpha$ is orthomodular, $[0,r\vee s]$ is
orthocomplemented, hence contains at least four atoms.
\end{remark}

\begin{theorem}\label{TheoremCoveringinTensorF}
Let $\{\l_\alpha\in\cCal(\s_\alpha)\}_{\alpha\in\Omega}$.
%
\begin{enumerate}
%
\item If each $\l_\alpha$ has the covering property and there is
at most one $\beta\in\Omega$ such that $\l_\beta\ne 2^{\s_\beta}$,
then $\otf_\alpha\l_\alpha$ has the covering property.
%
\item Suppose that each $\l_\alpha$ different from $2^{\s_\alpha}$
contains $\mo_4$. If $\otf_\alpha\l_\alpha$ has the covering
property, then there is at most one $\beta\in\Omega$ such that
$\l_\beta\ne 2^{\s_\beta}$.
%
\end{enumerate}
%
\end{theorem}

\begin{proof} (1) Let $\mathbf{\s}=\prod_\alpha\s_\alpha$,
$a\in\otf_\alpha\l_\alpha$, $q\in\mathbf{\s}$ not in $a$, and
$R=q\cup a$. By Lemma \ref{LemmaJoininTensorF},
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\[q\vee_\otfsub a=\vee_{\beta}R=\bigcup_{p\in\mathbf{\s}}p[\vee
R_{\beta}[p]]\, .\]
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
Now, by definition,
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\[R_{\beta}[p]=\pi_\beta(p[\s_\beta]\cap(q\cup a))
=q_\beta[p]\cup a_\beta[p]\, ,\]
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
and $q_{\beta}[p]=q_{\beta}$ if $p\in q[\s_{\beta}]$, and
$q_{\beta}[p]=\emptyset$ otherwise. Hence, for all $p$ not in
$q[\s_{\beta}]$, $R_{\beta}[p]=a_{\beta}[p]\in\l_{\beta}$, and if
$p\in q[\s_{\beta}]$, then $R_{\beta}[p]=q_{\beta}\cup
a_{\beta}[p]$. Therefore, we find that
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\[q\cup a\subseteq q\vee_\otfsub a=
q[q_{\beta}\vee a_{\beta}[q]]\bigcup_{p\not\in
q[\s_{\beta}]}p[a_{\beta}[p]] \subseteq a\cup q[q_{\beta}\vee
a_{\beta}[q]]\, .\]
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
As a consequence, $q\vee_{\otfsub} a\gtrdot a$.

(2) Let $\beta\ne\gamma\in\Omega$. Suppose that neither
$\l_\beta=2^{\s_\beta}$ nor $\l_\gamma=2^{\s_\gamma}$. Let
$p,\,q,\,r,\,s,\,t\in\mathbf{\s}$ such that
$p_\alpha=q_\alpha=r_\alpha=s_\alpha=t_\alpha$, for all $\alpha$
different from $\beta$ and $\gamma$, and such that
$t_\beta=p_\beta$ and $t_\gamma=q_\gamma$. Assume moreover that
for $\alpha=\beta$ and for $\alpha=\gamma$,
$p_\alpha,\,q_\alpha,\,r_\alpha$ and $s_\alpha$ are all different
and that $p_\alpha\vee q_\alpha$ covers
$p_\alpha,\,q_\alpha,\,r_\alpha$ and $s_\alpha$. By Definition
\ref{DefinitionSeparatedProduct}, $a=\{p,q,r\}$ and
$b=\{p,q,r,s\}$ are in $\otf_\alpha\l_\alpha$. Let $R^0=a\cup t$.
Then (see Lemma \ref{LemmaJoininTensorF}),
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\[\begin{split}R^1&:=\vee_\gamma R^0=R^0\cup p[p_\gamma\vee q_\gamma]\\
%
R^2&:=\vee_\beta R^1=R^0\cup p[p_\gamma\vee q_\gamma]\cup
q[p_\beta\vee q_\beta]\cup r[p_\beta\vee q_\beta]\cup
s[p_\beta\vee
q_\beta] \\
%
R^3&:=\vee_\gamma R^2=p[p_\beta\vee q_\beta,p_\gamma\vee
q_\gamma]\, .\end{split}\]
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
Hence, by Lemma \ref{LemmaJoininTensorF}, $a\vee_\otfsub t=R^3
\supsetneq b\supsetneq a$, therefore $\otf_\alpha\l_\alpha$ has
not the covering property.
\end{proof}

\begin{definition}
Let $\s_1,\,\s_2$ be sets, $\l_1=\mo_{\s_1}$, and
$\l_2=\mo_{\s_2}$ (see Definition \ref{DefinitionXi}). Then,
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\[\l_1\circ\l_2:=\l_1\aerts\l_2\cup\{R\in\Xi(\s_1\t\s_2)
\,;\,\abs R=3\}\, .\]
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\end{definition}

\begin{theorem}
Let $\s_1$ and $\s_2$ be sets, $\l_1=\mo_{\s_1}$,
$\l_2=\mo_{\s_2}$, $T=\aut(\l_1)\t\aut(\l_2)$, and
$\l\in\Sep_T(\l_1,\l_2)$. Then $\l$ has the covering property
$\Leftrightarrow$ $\l=\l_1\circ\l_2$.
\end{theorem}

\begin{proof} ($\Leftarrow$) Let $\mathbf{\s}=\s_1\t\s_2$,
$\Xi=\Xi(\s_1\t\s_2)$, and
$a\in\l_1\circ\l_2$. Then $a\in\Xi$ and $\abs{a}=3$, or
$a\in\l_1\aerts\l_2$. Hence one of the following cases holds.
%
\begin{enumerate}
%
\item $a\in\mathbf{\s}$.
%
\item $a\in\Xi$ and $\abs{a}=2$ or $3$.
%
\item $a=p_1\t\s_2$ or $a=\s_2\t p_2$ for some
$p=(p_1,p_2)\in\mathbf{\s}$.
%
\item $a=p_1\t\s_2\cup\s_1\t p_2$, for some $p\in\mathbf{\s}$
({\it i.e.} $a$ is a coatom).
%
\end{enumerate}
%
Hence, obviously $\l_1\circ\l_2$ has the covering property.

($\Rightarrow$) Let $\l\in\sep(\l_1,\l_2)$ with the covering
property, and $R\subseteq\mathbf{\s}$.

(1) By Lemma \ref{LemmaJoininTensorF} and Axiom P2, if
$R\not\in\Xi$ ({\it i.e.} there are $p^1,\,p^2\in R$ with
$p^1_i=p^2_i$ for $i=1$ or $2$), then $\vee R\in\l_1\aerts\l_2$.

(2) Suppose now that $R\in\Xi$. By Lemma
\ref{LemmaJoinP1diffQ1andP2diffQ2} part 1, if $\abs R\leq 2$, then
$R\in\l_1\aerts\l_2$, hence $R\in\l$. Moreover, if $\abs R\geq 3$,
then for all $s\in\mathbf{\s}$ not in $R$ with $s_1\in \pi_1(R)$
or $s_2\in\pi_2(R)$, we have $\vee (R\cup s)=1$.

(3) Suppose that $R\in\Xi$ and that $\abs R=3$. Write $a:=\vee R$.
{\bf Claim}: $a\ne 1$ [{\it Proof}\nobreak : Write $R=\{p,q,r\}$
and suppose that $a=1$. As we have seen $R^0:=\{p,q\}\in\l$. Hence
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\[1=a=r\vee R^0\supsetneq p_1\t\s_2\cup\s_1\t q_2\supsetneq R^0\, ,\]
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
a contradiction since $\l$ has the covering property.]

As a consequence, from part 2, $1\gtrdot a$. Moreover, $a\in\Xi$.
We write $a$ as $a=\{p_1\t f(p_1)\,;\,p_1\in\pi_1(a)\}$. Hence,
$f$ is injective.

(4) {\bf Claim}: $a=R$. [{\it Proof}\nobreak : First, note that
any bijection of $\s_2$ induces an automorphism of $\l_2$. Suppose
that $a\ne R$, hence $\abs{a}\geq 4$. Let $v_2\in\aut(\l_2)$ such
that its restriction to $\pi_2(a)$ is different from the identity,
and with at least three fixed points in $\pi_2(a)$.  By Axiom P4,
there is $u\in\aut(\l)$ such that on $\mathbf{\s}$, $u$ equals
$\mathsf{id}\t v_2$. Hence $c:=\{(p_1,v_2\circ f(p_1))\,;\,
p_1\in\pi_1(a)\}\in\l$ and $c\cap a\ne a$, therefore $1$ does not
cover $c\cap a$. Moreover $\abs{c\cap a}\geq 3$; whence a
contradiction by part 3.]

(5) Finally, suppose that $R\in\Xi$ and $\abs{R}\geq 4$. Let
$b=\vee R$. {\bf Claim}: $b=1$. [{\it proof}\nobreak : Let
$R_0\in\Xi$ with $R_0\subseteq R$ and $\abs{R_0}=3$. By what
precedes, $R_0\in\l$ and $1\gtrdot R_0$. As a consequence, $b=1$.]
\end{proof}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{An example with the covering
property}\label{SectionAnExample}

Let $\h_1$ and $\h_2$ be complex Hilbert spaces and
$T=\uni(\h_1)\t\uni(\h_2)$. In this section, we give an example in
$\Sep_T(\pro(\h_1),\pro(\h_2))$, denoted by
$\pro(\h_1)\ots\pro(\h_2)$, which has the covering property.
Moreover, $\pro(\h_1)\ots\pro(\h_2)$ is coatomistic, but, as
expected, the dual has not the covering property, {\it i.e.
$\pro(\h_1)\ots\pro(\h_2)$ is not a DAC-lattice.}

\begin{definition}\label{DefinitionOTS}
Let $\h_1$ and $\h_2$ be complex Hilbert spaces,
$\mathbf{\s}=\s_{\h_1}\t\s_{\h_2}$, $\l_1=\pro(\h_1)$,
$\l_2=\pro(\h_2)$, and $\s_\ot$ the set of one-dimensional
subspaces of $\h_1\ot\h_2$. For $V\in\pro(\h_1\ot\h_2)$, define
$\prs[V]:=\{(p_1,p_2)\in\mathbf{\s}\,;\,p_1\ot p_2\in V\}$. Then,
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\[\l_1\ots\l_2:=\{\prs[V]\,;\, V\in \pro(\h_1\ot \h_2)\}\, ,\]
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
ordered by set-inclusion. For $A\subseteq\mathbf{\s}$, we write
$A^\p:=\{q\in\s_\ot\,;\,\langle q,p_1\ot p_2\rangle=0,\,\forall
(p_1,p_2)\in A\}$, where $\langle-\vert-\rangle$ denotes the
scalar product in $\h_1\ot\h_2$. Moreover, we denote the set of
antilinear maps from $\h_1$ to $\h_2$ by $\mathcal{A}(\h_1,\h_2)$.
\end{definition}

\begin{proposition} Let $m$ and $n$ be
integers, $\h_1=\mathbb{C}^m$, $\h_2=\mathbb{C}^n$,
$\l_1=\pro(\h_1)$, $\l_2=\pro(\h_2)$, and
$\mathbf{\s}=\s_{\h_1}\t\s_{\h_2}$. For $A\in
\mathcal{A}(\h_1,\h_2)$, define $X_A\subseteq\mathbf{\s}$ as
$X_A:=\cup\{p_1\t(A(p_1)^{\p_2})\,;\,p_1\in \s_1\}$. Then,
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\[\l_1\ots\l_2=\{\cap\omega\,;\,\omega\subseteq\{X_A\,;\,A\in
\mathcal{A}(\h_1,\h_2)\}\}\, .\]
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\end{proposition}

\begin{proof}
Let $\{e_i^1\}_{1\leq i\leq m}$ and $\{e_j^2\}_{1\leq j\leq n}$
denote the canonical basis of $\mathbb{C}^m$ and $\mathbb{C}^n$
respectively.

Let $v\in\mathbb{C}^m\ot\mathbb{C}^n$ and
$p=(p_1,p_2)\in\mathbf{\s}$ with $p_1=\mathbb{C}w_1$ and
$p_2=\mathbb{C}w_2$. Write $v$, $w_1$ and $w_2$ as
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\[v=\sum_{i=1}^m\sum_{j=1}^n s_{ij} e_i^1\ot e_j^2,\
w_1=\sum_{i=1}^m\lambda_i e_i^1,\ \mbox{and}\ w_2=\sum_{j=1}^n
\mu_j e_j^2\, .\]
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
Let
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\[\lambda=(\lambda_1,\cdots,\lambda_m)^\top\ \mbox{and}\
\mu=(\mu_1,\cdots,\mu_n)^\top\, .\]
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
Let $S$ be the $m\t n$ matrix defined as $S_{ij}=s_{ij}$. Then
$p\in \prs[v^\p]$ if and only if $\langle w_1\ot w_2,v\rangle=0$,
hence if and only if $\overline{\mu}^\top S^\top
\overline{\lambda}=0$. Let $A$ be the antilinear map defined by
the matrix $S^\top$. Then, $p\in \prs[v^\p]$ if and only if
$p_2\in A(p_1)^\p$, that is if and only if $p\in X_A$. As a
consequence, $\prs[v^\p]=X_A$.

On the other hand, if $A\in \mathcal{A}(\h_1,\h_2)$, then
$X_A=\prs[v^\p]$, where $v$ is given by the formula above with
$s_{ij}=(A^\top)_{ij}$.
\end{proof}

\begin{remark}
Let $\h$ be a complex Hilbert space of dimension $\geq 3$. Then,
by Wigner's theorem (see \cite{Faure/Froelicher:handbook}, Theorem
14.3.6), any ortho-automorphism of $\pro(\h)$ is induced by a
unitary or antiunitary map on $\h$. Note that if $v_1$ is a
unitary map on $\h_1$ and $v_2$ is an antiunitary map on $\h_2$,
then $v=v_1\t v_2$ does not induce an automorphism of
$\l_1\ots\l_2$. Indeed, let $X_A$ be a coatom. Then
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\[v(X_A)=\bigcup_{p_1\in\s_1}v_1(p_1)\t
v_2(A(p_1)^{\p_2})=\bigcup_{p_1\in\s_1}p_1\t((v_2\circ A\circ
v_1^{-1}(p_1))^{\p_2})\, .\]
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
Now, since $A$ and $v_2$ are antilinear and $v_1$ is linear, it
follows that $v_2\circ A\circ v_1^{-1}$ is linear, hence $v(X_A)$
is not a coatom of $\l_1\ots\l_2$.
\end{remark}

\begin{theorem}
Let $\h_1$ and $\h_2$ be complex Hilbert spaces,
$\mathbf{\s}=\s_{\h_1}\t\s_{\h_2}$, $\l_1=\pro(\h_1)$,
$\l_2=\pro(\h_2)$, and $T=\uni(\h_1)\t\uni(\h_2)$. Then
\begin{enumerate}
\item for all $A\subseteq\mathbf{\s}$, we have $\vee_\otssub
A=\prs[A^{\p\p}]$,
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\item $\l_1\ots\l_2\in\Sep_T(\l_1,\l_2)$,
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\item $\l_1\ots\l_2$ has the covering property and is
coato\-mistic, but, if $\l_1\ne 2\ne\l_2$ ({\it i.e.} the
dimension of $\h_1$ and $\h_2$ is $\geq 2$), the dual has not the
covering property,
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\item $\l_1\aerts\l_2=\{\prs[V]\,;\, V\in\pro(\h_1\ot \h_2),\,
V=\prs[V]^{\p\p}, V^\p=\prs[V^\p]^{\p\p}\}$ ({\it i.e.} both $V$
and $V^\p$ are spanned by product vectors),
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\item if $\l_1\ne 2\ne\l_2$, then $\l_1\aerts\l_2\varsubsetneqq
\l_1\ots\l_2\varsubsetneqq\l_1\otf\l_2$.
\end{enumerate}
\end{theorem}

\begin{proof} (1) Follows directly from Definition \ref{DefinitionOTS}.

(2) Obviously, $\l_1\ots\l_2\in\cCal(\mathbf{\s})$. Let
$a\in\l_1\aerts\l_2$. By Definition \ref{DefinitionOrthAerts},
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\[a^\#=\prs[a^\p]\, .\]
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
Hence, $a^\#\subseteq a^\p$, $a^{\p\p}\subseteq a^{\#\p}$, thus
$\prs[a^{\p\p}]\subseteq a^{\#\#}=a$, therefore
$a=\prs[a^{\p\p}]$. As a consequence,
$\l_1\aerts\l_2\subseteq\l_1\ots\l_2$.

Let $V\in\pro(\h_1\ot\h_2)$ such that $\prs[V]=p_1\t A_2$. Then
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\[p_1\ot(A_2^{\p_2\p_2})=(p_1\ot A_2)^{\p\p}\subseteq V\, ,\]
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
therefore $A_2^{\p_2\p_2}\subseteq A_2$, hence $A_2\in\l_2$. As a
consequence, Axiom P3 holds.

Axiom P4 with $T=\uni(\h_1)\t\uni(\h_2)$ holds in
$\pro(\h_1\ot\h_2)$, therefore obviously also in $\l_1\ots\l_2$.

(3) The covering property holds in $\pro(\h_1\ot \h_2)$ (see
\cite{Maeda/Maeda:handbook}, Theorem 34.2), hence, by part 1, also
in $\l_1\ots\l_2$. Moreover, since $\pro(\h_1\ot\h_2)$ is
coatomistic, so is $\l_1\ots\l_2$.

Finally, let $p\in\mathbf{\s}$. Then $x=\prs[p^\p]=p^{\#}$ is
coatom of $\l_1\ots \l_2$. Now, there is $R\in\Xi(\mathbf{\s})$
(see Definition \ref{DefinitionXi}) with $\abs R=2$, such that
$x\cap R=\emptyset$. By Lemma \ref{LemmaJoinP1diffQ1andP2diffQ2},
$R\in\l_1\ots\l_2$. Moreover $x\vee_\otssub R=1$ since $x$ is a
coatom. Hence, writing $R=\{p,q\}$, and the order relation, meet,
join, bottom and top elements in $(\l_1\ots\l_2)^*$ by $\leq_*$,
$\land_*$, $\vee_*$, $0_*$, and $1_*$ respectively, we find that
$x\land_* R=0_*$ and $x\vee_* R=1_*\gneq_* p\gneq_* R$. Therefore,
$(\l_1\ots\l_2)^*$ has not the covering property.

(4) Let $p\in\mathbf{\s}$ and $q\in p^{\#\p}$. Write
$q=\mathbb{C}v$ with $v\in\h_1\ot\h_2$. For $i=1$ and $i=2$, let
$\{w_i^k\}$ be an ortho-basis of $p_i^{\p_i}$, and let $x_i\in
p_i$ ({\it i.e.} $p_i=\mathbb{C}x_i$). Then $v$ can be decomposed
as
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\[v=\alpha x_1\ot x_2+\sum_{k_2}\beta_{k_2}x_1\ot
w_2^{k_2}+\sum_{k_1}\beta_{k_1}w_1^{k_1}\ot
x_2+\sum_{l_1l_2}\gamma_{l_1l_2}w_1^{l_1}\ot w_2^{l_2}\, .\]
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
Now, $p^\#=p_1^{\p_1}\t\s_{\h_2}\cup\s_{\h_1}\t p_2^{\p_2}$.
Hence, since $v\in p^{\#\p}$, we find that
$\beta_{k_1}=\beta_{k_2}=\gamma_{l_1l_2}=0$, for all
$k_1,\,k_2,\,l_1$ and $l_2$. Therefore, $v\in p$, hence
$p^{\#\p}=p$.

Let $a\in\l_1\aerts\l_2$. From part 1, $a=\prs[a^{\p\p}]$. On the
other hand, $a^\#=\prs[a^\p]$. Now, $a=a^{\#\#}=\cap\{p^\#\,;\,
p\in a^\#\}$. Hence, by what precedes,
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\[a^\p=\left( \cup\{ p^{\#\p}\,;\, p\in a^\#\}\right )^{\p\p}=
a^{\#\p\p}\, ,\]
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
therefore $a^\p$ is also spanned by product vectors.

Let $V\in\pro(\h_1\ot \h_2)$ such that both $V$ and $V^\p$ are
spanned by product vectors. Let $a:=\prs[V]$ and $b:=\prs[V^\p]$.
Since $\prs[V]^{\p\p}=V$ and $\prs[V^\p]^{\p\p}=V^\p$, we find
that
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\[\prs[\prs[V]^\p]= \prs[V^\p]\  \mbox{and}\ \prs[\prs[V^\p]
^\p]=\prs[V]\, .\]
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
Therefore $a^\#=b$ and $b^\#=a$. As a consequence,
$a\in\l_1\aerts\l_2$.

(5) By part 4, $\l_1\aerts\l_2\ne\l_1\ots\l_2$. On the other hand,
by Theorem \ref{TheoremCoveringinTensorF}, $\l_1\otf\l_2$ has not
the covering property, whereas by part 3, $\l_1\ots\l_2$ has the
covering property. As a consequence,
$\l_1\ots\l_2\ne\l_1\otf\l_2$.\end{proof}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section*{Acknowledgments}
An important part of this work was done during a stay at McGill.
In this connection, I would like to thanks M. Barr for his
hospitality.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%\bibliographystyle{amsplain}
\bibliographystyle{abbrv}
\bibliography{SLTPinQL3}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%\sto
\end{document}

diagxy.tex
 \def\dated#1{\def\thedate{#1}}
 \dated{2004-03-08}
 % This is a package of commutative diagram macros built on top of Xy-pic
 % by Michael Barr (email:  barr@barrs.org).  Its use is unrestricted.  It
 % may be freely distributed, unchanged, for non-commercial or commercial
 % use.  If changed, it must be renamed.  Inclusion in a commercial
 % software package is also permitted, but I would appreciate receiving a
 % free copy for my personal examination and use.  There are no guarantees
 % that this package is good for anything.  I have tested it with LaTeX 2e,
 % LaTeX 2.09 and Plain TeX.  Although I know of no reason it will not work
 % with AMSTeX, I have not tested it.
 % Added 2003-05-10: I now know that the only clash is with \square in
 % amssymb, which should therefore be loaded first.  If the amssymb
 % \square is required, you can \let\box\square and use \box instead.

\newcount\atcode \atcode=\catcode`\@%
\catcode`\@=12
\input xy
\xyoption{arrow}
\xyoption{curve}

\newdir{ >}{{ }*!/-.9em/@{>}}
\newdir{ (}{{ }*!/-.5em/@{(}}
\newdir^{ (}{{ }*!/-.5em/@^{(}}
\newdir{< }{!/.9em/@{<}*{ }}

\newdimen\high%
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\newcount\deltaX%
\newcount\deltaY%

\newdimen\wdth
\newcount\xend%
\newcount\yend%
\newcount\Xend
\newcount\Yend
\newcount\xpos%
\newcount\ypos%
\newcount\default \default=500%
\newcount\defaultmargin \defaultmargin=150
\newcount\topw%
\newcount\botw%
\newcount\Xpos%
\newcount\Ypos%
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\catcode`\@=11
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\def\ifnextchar#1#2#3{\let\@tempe
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\def\@ifnch{\ifx \@tempc \@sptoken \let\@tempd\@xifnch
      \else \ifx \@tempc \@tempe\let\@tempd\@tempa\else\let\@tempd\@tempb\fi
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\def\:{\let\@sptoken= } \:  % this makes \@sptoken a space token
\def\:{\@xifnch} \expandafter\def\: {\futurelet\@tempc\@ifnch}%
\else
\let\ifnextchar\@ifnextchar
\fi
\ifx\check@mathfonts\undefined
\else \check@mathfonts
\fi
\newdimen\axis \axis=\fontdimen22\textfont2
\ul=.01em%
\X@xbase =.01em%
\Y@ybase =.01em%
\def\scalefactor#1{\ul=#1\ul \X@xbase=#1\X@xbase \Y@ybase=#1\Y@ybase}%
\catcode`\@=12%

\def\fontscale#1{%
\if#1h\relax
\font\xydashfont=xydash10 scaled \magstephalf
\font\xyatipfont=xyatip10 scaled \magstephalf
\font\xybtipfont=xybtip10 scaled \magstephalf
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\font\xybtipfont=xybtip10 scaled \magstep#1%
\font\xybsqlfont=xybsql10 scaled \magstep#1%
\font\xycircfont=xycirc10 scaled \magstep#1%
\fi}

\def\bfig{\vcenter\bgroup\xy}
\def\efig{\endxy\egroup}

\def\car#1#2\nil{#1}%

\def\morphism{\ifnextchar({\morphismp}{\morphismp(0,0)}}%
\def\morphismp(#1){\ifnextchar|{\morphismpp(#1)}{\morphismpp(#1)|a|}}%
\def\morphismpp(#1)|#2|{\ifnextchar/{\morphismppp(#1)|#2|}%
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\def\morphismpppp(#1,#2)|#3|/#4/<#5,#6>[#7`#8;#9]{%
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\yend#2\advance \yend by #6%
\domorphism(#1,#2)|#3|/#4/<#5,#6>[{#7}`{#8};{#9}]}


\def\domorphism(#1,#2)|#3|/#4/<#5,#6>[#7`#8;#9]{%
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\else%
 \if#3l%
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     \POS(#1,#2)*+!!<0ex,\axis>{#7}\ar@{#4} (\xend,\yend)*+!!<0ex,\axis>{#8}%
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     \POS(#1,#2)*+!!<0ex,\axis>{#7}\ar@{#4}|-*+<1pt,4pt>{\labelstyle#9}
         (\xend,\yend)*+!!<0ex,\axis>{#8}%
   \fi
 \else \if#3r%
  \ifnum #6<0%
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  \else%
   \POS(#1,#2)*+!!<0ex,\axis>{#7}\ar@{#4}_-{#9} (\xend,\yend)*+!!<0ex,\axis>{#8}%
  \fi%
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  \ifnum #5>0%
   \POS(#1,#2)*+!!<0ex,\axis>{#7}\ar@{#4}^-{#9} (\xend,\yend)*+!!<0ex,\axis>{#8}%
  \else%
   \POS(#1,#2)*+!!<0ex,\axis>{#7}\ar@{#4}_-{#9} (\xend,\yend)*+!!<0ex,\axis>{#8}%
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\fi\ignorespaces}%

\def\vector(#1,#2)/#3/<#4,#5>{%
 \xend#1 \yend#2 \advance\xend by #4 \advance\yend by #5
     \POS(#1,#2)\ar#3 (\xend,\yend)}

\def\squarepppp(#1,#2)|#3|/#4`#5`#6`#7/<#8>[#9]{%
\xpos#1\ypos#2%
\def\next|##1##2##3##4|{%
 \def\xa{##1}\def\xb{##2}\def\xc{##3}\def\xd{##4}\ignorespaces}%
\next|#3|%
\def\next<##1,##2>{\deltax=##1\deltay=##2\ignorespaces}%
\next<#8>%
\def\next[##1`##2`##3`##4;##5`##6`##7`##8]{%
    \def\nodea{##1}\def\nodeb{##2}\def\nodec{##3}\def\noded{##4}%
    \def\labela{##5}\def\labelb{##6}\def\labelc{##7}\def\labeld{##8}\ignorespaces}%
\next[#9]%
\morphism(\xpos,\ypos)|\xd|/{#7}/<\deltax,0>[\nodec`\noded;\labeld]%
\advance \ypos by \deltay%
\morphism(\xpos,\ypos)|\xb|/{#5}/<0,-\deltay>[\nodea`\nodec;\labelb]%
\morphism(\xpos,\ypos)|\xa|/{#4}/<\deltax,0>[\nodea`\nodeb;\labela]%
 \advance \xpos by \deltax%
\morphism(\xpos,\ypos)|\xc|/{#6}/<0,-\deltay>[\nodeb`\noded;\labelc]%
\ignorespaces}%

\def\square{\ifnextchar({\squarep}{\squarep(0,0)}}%
\def\squarep(#1){\ifnextchar|{\squarepp(#1)}{\squarepp(#1)|alrb|}}%
\def\squarepp(#1)|#2|{\ifnextchar/{\squareppp(#1)|#2|}%
    {\squareppp(#1)|#2|/>`>`>`>/}}%
\def\squareppp(#1)|#2|/#3`#4`#5`#6/{%
    \ifnextchar<{\squarepppp(#1)|#2|/#3`#4`#5`#6/}%
    {\squarepppp(#1)|#2|/#3`#4`#5`#6/<\default,\default>}}%

\def\ptrianglepppp(#1,#2)|#3|/#4`#5`#6/<#7>[#8]{%
\xpos#1\ypos#2%
\def\next|##1##2##3|{\def\xa{##1}\def\xb{##2}\def\xc{##3}}%
\next|#3|%
\def\next<##1,##2>{\deltax=##1\deltay=##2\ignorespaces}%
\next<#7>%
\def\next[##1`##2`##3;##4`##5`##6]{%
    \def\nodea{##1}\def\nodeb{##2}\def\nodec{##3}%
    \def\labela{##4}\def\labelb{##5}\def\labelc{##6}}%
\next[#8]%
\advance\ypos by \deltay%
\morphism(\xpos,\ypos)|\xa|/{#4}/<\deltax,0>[\nodea`\nodeb;\labela]%
\morphism(\xpos,\ypos)|\xb|/{#5}/<0,-\deltay>[\nodea`\nodec;\labelb]%
\advance\xpos by \deltax%
\morphism(\xpos,\ypos)|\xc|/{#6}/<-\deltax,-\deltay>[\nodeb`\nodec;\labelc]%
\ignorespaces}%

\def\qtrianglepppp(#1,#2)|#3|/#4`#5`#6/<#7>[#8]{%
\xpos#1\ypos#2%
\def\next|##1##2##3|{\def\xa{##1}\def\xb{##2}\def\xc{##3}}%
\next|#3|%
\def\next<##1,##2>{\deltax=##1\deltay=##2\ignorespaces}%
\next<#7>%
\def\next[##1`##2`##3;##4`##5`##6]{%
    \def\nodea{##1}\def\nodeb{##2}\def\nodec{##3}%
    \def\labela{##4}\def\labelb{##5}\def\labelc{##6}}%
\next[#8]%
\advance\ypos by \deltay%
\morphism(\xpos,\ypos)|\xa|/{#4}/<\deltax,0>[\nodea`\nodeb;\labela]%
\morphism(\xpos,\ypos)|\xb|/{#5}/<\deltax,-\deltay>[\nodea`\nodec;\labelb]%
\advance\xpos by \deltax%
\morphism(\xpos,\ypos)|\xc|/{#6}/<0,-\deltay>[\nodeb`\nodec;\labelc]%
\ignorespaces}%

\def\dtrianglepppp(#1,#2)|#3|/#4`#5`#6/<#7>[#8]{%
\xpos#1\ypos#2%
\def\next|##1##2##3|{\def\xa{##1}\def\xb{##2}\def\xc{##3}}%
\next|#3|%
\def\next<##1,##2>{\deltax=##1\deltay=##2\ignorespaces}%
\next<#7>%
\def\next[##1`##2`##3;##4`##5`##6]{%
    \def\nodea{##1}\def\nodeb{##2}\def\nodec{##3}%
    \def\labela{##4}\def\labelb{##5}\def\labelc{##6}}%
\next[#8]%
\morphism(\xpos,\ypos)|\xc|/{#6}/<\deltax,0>[\nodeb`\nodec;\labelc]%
\advance\ypos by \deltay\advance \xpos by \deltax%
\morphism(\xpos,\ypos)|\xa|/{#4}/<-\deltax,-\deltay>[\nodea`\nodeb;\labela]%
\morphism(\xpos,\ypos)|\xb|/{#5}/<0,-\deltay>[\nodea`\nodec;\labelb]%
\ignorespaces}%

\def\btrianglepppp(#1,#2)|#3|/#4`#5`#6/<#7>[#8]{%
\xpos#1\ypos#2%
\def\next|##1##2##3|{\def\xa{##1}\def\xb{##2}\def\xc{##3}}%
\next|#3|%
\def\next<##1,##2>{\deltax=##1\deltay=##2\ignorespaces}%
\next<#7>%
\def\next[##1`##2`##3;##4`##5`##6]{%
    \def\nodea{##1}\def\nodeb{##2}\def\nodec{##3}%
    \def\labela{##4}\def\labelb{##5}\def\labelc{##6}}%
\next[#8]%
\morphism(\xpos,\ypos)|\xc|/{#6}/<\deltax,0>[\nodeb`\nodec;\labelc]%
\advance\ypos by \deltay%
\morphism(\xpos,\ypos)|\xa|/{#4}/<0,-\deltay>[\nodea`\nodeb;\labela]%
\morphism(\xpos,\ypos)|\xb|/{#5}/<\deltax,-\deltay>[\nodea`\nodec;\labelb]%
\ignorespaces}%

\def\Atrianglepppp(#1,#2)|#3|/#4`#5`#6/<#7>[#8]{%
\xpos#1\ypos#2%
\def\next|##1##2##3|{\def\xa{##1}\def\xb{##2}\def\xc{##3}}%
\next|#3|%
\def\next<##1,##2>{\deltax=##1\deltay=##2\ignorespaces}%
\next<#7>%
\def\next[##1`##2`##3;##4`##5`##6]{%
    \def\nodea{##1}\def\nodeb{##2}\def\nodec{##3}%
    \def\labela{##4}\def\labelb{##5}\def\labelc{##6}}%
\next[#8]%
\multiply\deltax by 2%
\morphism(\xpos,\ypos)|\xc|/{#6}/<\deltax,0>[\nodeb`\nodec;\labelc]%
\divide\deltax by 2
\advance\ypos by \deltay\advance\xpos by \deltax%
\morphism(\xpos,\ypos)|\xa|/{#4}/<-\deltax,-\deltay>[\nodea`\nodeb;\labela]%
\morphism(\xpos,\ypos)|\xb|/{#5}/<\deltax,-\deltay>[\nodea`\nodec;\labelb]%
\ignorespaces}%

\def\Vtrianglepppp(#1,#2)|#3|/#4`#5`#6/<#7>[#8]{%
\xpos#1\ypos#2%
\def\next|##1##2##3|{\def\xa{##1}\def\xb{##2}\def\xc{##3}}%
\next|#3|%
\def\next<##1,##2>{\deltax=##1\deltay=##2\ignorespaces}%
\next<#7>%
\def\next[##1`##2`##3;##4`##5`##6]{%
    \def\nodea{##1}\def\nodeb{##2}\def\nodec{##3}%
    \def\labela{##4}\def\labelb{##5}\def\labelc{##6}}%
\next[#8]%
\advance\ypos by \deltay%
\morphism(\xpos,\ypos)|\xb|/{#5}/<\deltax,-\deltay>[\nodea`\nodec;\labelb]%
\multiply\deltax by 2%
\morphism(\xpos,\ypos)|\xa|/{#4}/<\deltax,0>[\nodea`\nodeb;\labela]%
\advance\xpos by \deltax \divide \deltax by 2
\morphism(\xpos,\ypos)|\xc|/{#6}/<-\deltax,-\deltay>[\nodeb`\nodec;\labelc]%
\ignorespaces}%

\def\Ctrianglepppp(#1,#2)|#3|/#4`#5`#6/<#7>[#8]{%
\xpos#1\ypos#2%
\def\next|##1##2##3|{\def\xa{##1}\def\xb{##2}\def\xc{##3}}%
\next|#3|%
\def\next<##1,##2>{\deltax=##1\deltay=##2\ignorespaces}%
\next<#7>%
\def\next[##1`##2`##3;##4`##5`##6]{%
    \def\nodea{##1}\def\nodeb{##2}\def\nodec{##3}%
    \def\labela{##4}\def\labelb{##5}\def\labelc{##6}}%
\next[#8]%
\advance \ypos by \deltay%
\morphism(\xpos,\ypos)|\xc|/{#6}/<\deltax,-\deltay>[\nodeb`\nodec;\labelc]%
\advance\ypos by \deltay \advance \xpos by \deltax%
\morphism(\xpos,\ypos)|\xa|/{#4}/<-\deltax,-\deltay>[\nodea`\nodeb;\labela]%
\multiply\deltay by 2%
\morphism(\xpos,\ypos)|\xb|/{#5}/<0,-\deltay>[\nodea`\nodec;\labelb]%
\ignorespaces}%

\def\Dtrianglepppp(#1,#2)|#3|/#4`#5`#6/<#7>[#8]{%
\xpos#1\ypos#2%
\def\next|##1##2##3|{\def\xa{##1}\def\xb{##2}\def\xc{##3}}%
\next|#3|%
\def\next<##1,##2>{\deltax=##1\deltay=##2\ignorespaces}%
\next<#7>%
\def\next[##1`##2`##3;##4`##5`##6]{%
    \def\nodea{##1}\def\nodeb{##2}\def\nodec{##3}%
    \def\labela{##4}\def\labelb{##5}\def\labelc{##6}}%
\next[#8]%
\advance\xpos by \deltax \advance\ypos by \deltay%
\morphism(\xpos,\ypos)|\xc|/{#6}/<-\deltax,-\deltay>[\nodeb`\nodec;\labelc]%
\advance\xpos by -\deltax \advance\ypos by \deltay%
\morphism(\xpos,\ypos)|\xb|/{#5}/<\deltax,-\deltay>[\nodea`\nodeb;\labelb]%
\multiply \deltay by 2%
\morphism(\xpos,\ypos)|\xa|/{#4}/<0,-\deltay>[\nodea`\nodec;\labela]%
\ignorespaces}%

\def\ptriangle{\ifnextchar({\ptrianglep}{\ptrianglep(0,0)}}%
\def\ptrianglep(#1){\ifnextchar|{\ptrianglepp(#1)}{\ptrianglepp(#1)|alr|}}%
\def\ptrianglepp(#1)|#2|{\ifnextchar/{\ptriangleppp(#1)|#2|}%
    {\ptriangleppp(#1)|#2|/>`>`>/}}%
\def\ptriangleppp(#1)|#2|/#3`#4`#5/{%
    \ifnextchar<{\ptrianglepppp(#1)|#2|/#3`#4`#5/}%
    {\ptrianglepppp(#1)|#2|/#3`#4`#5/<\default,\default>}}%

\def\qtriangle{\ifnextchar({\qtrianglep}{\qtrianglep(0,0)}}%
\def\qtrianglep(#1){\ifnextchar|{\qtrianglepp(#1)}{\qtrianglepp(#1)|alr|}}%
\def\qtrianglepp(#1)|#2|{\ifnextchar/{\qtriangleppp(#1)|#2|}%
    {\qtriangleppp(#1)|#2|/>`>`>/}}%
\def\qtriangleppp(#1)|#2|/#3`#4`#5/{%
    \ifnextchar<{\qtrianglepppp(#1)|#2|/#3`#4`#5/}%
    {\qtrianglepppp(#1)|#2|/#3`#4`#5/<\default,\default>}}%

\def\dtriangle{\ifnextchar({\dtrianglep}{\dtrianglep(0,0)}}%
\def\dtrianglep(#1){\ifnextchar|{\dtrianglepp(#1)}{\dtrianglepp(#1)|lrb|}}%
\def\dtrianglepp(#1)|#2|{\ifnextchar/{\dtriangleppp(#1)|#2|}%
    {\dtriangleppp(#1)|#2|/>`>`>/}}%
\def\dtriangleppp(#1)|#2|/#3`#4`#5/{%
    \ifnextchar<{\dtrianglepppp(#1)|#2|/#3`#4`#5/}%
    {\dtrianglepppp(#1)|#2|/#3`#4`#5/<\default,\default>}}%

\def\btriangle{\ifnextchar({\btrianglep}{\btrianglep(0,0)}}%
\def\btrianglep(#1){\ifnextchar|{\btrianglepp(#1)}{\btrianglepp(#1)|lrb|}}%
\def\btrianglepp(#1)|#2|{\ifnextchar/{\btriangleppp(#1)|#2|}%
    {\btriangleppp(#1)|#2|/>`>`>/}}%
\def\btriangleppp(#1)|#2|/#3`#4`#5/{%
    \ifnextchar<{\btrianglepppp(#1)|#2|/#3`#4`#5/}%
    {\btrianglepppp(#1)|#2|/#3`#4`#5/<\default,\default>}}%

\def\Atriangle{\ifnextchar({\Atrianglep}{\Atrianglep(0,0)}}%
\def\Atrianglep(#1){\ifnextchar|{\Atrianglepp(#1)}{\Atrianglepp(#1)|lrb|}}%
\def\Atrianglepp(#1)|#2|{\ifnextchar/{\Atriangleppp(#1)|#2|}%
    {\Atriangleppp(#1)|#2|/>`>`>/}}%
\def\Atriangleppp(#1)|#2|/#3`#4`#5/{%
    \ifnextchar<{\Atrianglepppp(#1)|#2|/#3`#4`#5/}%
    {\Atrianglepppp(#1)|#2|/#3`#4`#5/<\default,\default>}}%

\def\Vtriangle{\ifnextchar({\Vtrianglep}{\Vtrianglep(0,0)}}%
\def\Vtrianglep(#1){\ifnextchar|{\Vtrianglepp(#1)}{\Vtrianglepp(#1)|alb|}}%
\def\Vtrianglepp(#1)|#2|{\ifnextchar/{\Vtriangleppp(#1)|#2|}%
    {\Vtriangleppp(#1)|#2|/>`>`>/}}%
\def\Vtriangleppp(#1)|#2|/#3`#4`#5/{%
    \ifnextchar<{\Vtrianglepppp(#1)|#2|/#3`#4`#5/}%
    {\Vtrianglepppp(#1)|#2|/#3`#4`#5/<\default,\default>}}%

\def\Ctriangle{\ifnextchar({\Ctrianglep}{\Ctrianglep(0,0)}}%
\def\Ctrianglep(#1){\ifnextchar|{\Ctrianglepp(#1)}{\Ctrianglepp(#1)|arb|}}%
\def\Ctrianglepp(#1)|#2|{\ifnextchar/{\Ctriangleppp(#1)|#2|}%
    {\Ctriangleppp(#1)|#2|/>`>`>/}}%
\def\Ctriangleppp(#1)|#2|/#3`#4`#5/{%
    \ifnextchar<{\Ctrianglepppp(#1)|#2|/#3`#4`#5/}%
    {\Ctrianglepppp(#1)|#2|/#3`#4`#5/<\default,\default>}}%

\def\Dtriangle{\ifnextchar({\Dtrianglep}{\Dtrianglep(0,0)}}%
\def\Dtrianglep(#1){\ifnextchar|{\Dtrianglepp(#1)}{\Dtrianglepp(#1)|alb|}}%
\def\Dtrianglepp(#1)|#2|{\ifnextchar/{\Dtriangleppp(#1)|#2|}%
    {\Dtriangleppp(#1)|#2|/>`>`>/}}%
\def\Dtriangleppp(#1)|#2|/#3`#4`#5/{%
    \ifnextchar<{\Dtrianglepppp(#1)|#2|/#3`#4`#5/}%
    {\Dtrianglepppp(#1)|#2|/#3`#4`#5/<\default,\default>}}%


\def\Atrianglepairpppp(#1)|#2|/#3`#4`#5`#6`#7/<#8>[#9]{%
\def\next(##1,##2){\xpos##1\ypos##2}%
\next(#1)%
\def\next|##1##2##3##4##5|{\def\xa{##1}\def\xb{##2}%
\def\xc{##3}\def\xd{##4}\def\xe{##5}}%
\next|#2|%
\def\next<##1,##2>{\deltax=##1\deltay=##2\ignorespaces}%
\next<#8>%
\def\next[##1`##2`##3`##4;##5`##6`##7`##8`##9]{%
 \def\nodea{##1}\def\nodeb{##2}\def\nodec{##3}\def\noded{##4}%
 \def\labela{##5}\def\labelb{##6}\def\labelc{##7}\def\labeld{##8}\def\labele{##9}}%
\next[#9]%
\morphism(\xpos,\ypos)|\xd|/{#6}/<\deltax,0>[\nodeb`\nodec;\labeld]%
\advance\xpos by \deltax%
\morphism(\xpos,\ypos)|\xe|/{#7}/<\deltax,0>[\nodec`\noded;\labele]%
\advance\ypos by \deltay%
\morphism(\xpos,\ypos)|\xa|/{#3}/<-\deltax,-\deltay>[\nodea`\nodeb;\labela]%
\morphism(\xpos,\ypos)|\xb|/{#4}/<0,-\deltay>[\nodea`\nodec;\labelb]%
\morphism(\xpos,\ypos)|\xc|/{#5}/<\deltax,-\deltay>[\nodea`\noded;\labelc]%
\ignorespaces}%

\def\Vtrianglepairpppp(#1)|#2|/#3`#4`#5`#6`#7/<#8>[#9]{%
\def\next(##1,##2){\xpos##1\ypos##2}%
\next(#1)%
\def\next|##1##2##3##4##5|{\def\xa{##1}\def\xb{##2}%
\def\xc{##3}\def\xd{##4}\def\xe{##5}}%
\next|#2|%
\def\next<##1,##2>{\deltax=##1\deltay=##2\ignorespaces}%
\next<#8>%
\def\next[##1`##2`##3`##4;##5`##6`##7`##8`##9]{%
 \def\nodea{##1}\def\nodeb{##2}\def\nodec{##3}\def\noded{##4}%
 \def\labela{##5}\def\labelb{##6}\def\labelc{##7}\def\labeld{##8}\def\labele{##9}}%
\next[#9]%
\advance\ypos by \deltay%
\morphism(\xpos,\ypos)|\xa|/{#3}/<\deltax,0>[\nodea`\nodeb;\labela]%
\morphism(\xpos,\ypos)|\xc|/{#5}/<\deltax,-\deltay>[\nodea`\noded;\labelc]%
\advance\xpos by \deltax%
\morphism(\xpos,\ypos)|\xb|/{#4}/<\deltax,0>[\nodeb`\nodec;\labelb]%
\morphism(\xpos,\ypos)|\xd|/{#6}/<0,-\deltay>[\nodeb`\noded;\labeld]%
\advance\xpos by \deltax%
\morphism(\xpos,\ypos)|\xe|/{#7}/<-\deltax,-\deltay>[\nodec`\noded;\labele]%
\ignorespaces}%

\def\Ctrianglepairpppp(#1)|#2|/#3`#4`#5`#6`#7/<#8>[#9]{%
\def\next(##1,##2){\xpos##1\ypos##2}%
\next(#1)%
\def\next|##1##2##3##4##5|{\def\xa{##1}\def\xb{##2}%
\def\xc{##3}\def\xd{##4}\def\xe{##5}}%
\next|#2|%
\def\next<##1,##2>{\deltax=##1\deltay=##2\ignorespaces}%
\next<#8>%
\def\next[##1`##2`##3`##4;##5`##6`##7`##8`##9]{%
 \def\nodea{##1}\def\nodeb{##2}\def\nodec{##3}\def\noded{##4}%
 \def\labela{##5}\def\labelb{##6}\def\labelc{##7}\def\labeld{##8}\def\labele{##9}}%
\next[#9]%
\advance\ypos by \deltay%
\morphism(\xpos,\ypos)|\xe|/{#7}/<0,-\deltay>[\nodec`\noded;\labele]%
\advance\xpos by -\deltax%
\morphism(\xpos,\ypos)|\xc|/{#5}/<\deltax,0>[\nodeb`\nodec;\labelc]%
\morphism(\xpos,\ypos)|\xd|/{#6}/<\deltax,-\deltay>[\nodeb`\noded;\labeld]%
\advance\ypos by \deltay%
\advance\xpos by \deltax%
\morphism(\xpos,\ypos)|\xa|/{#3}/<-\deltax,-\deltay>[\nodea`\nodeb;\labela]%
\morphism(\xpos,\ypos)|\xb|/{#4}/<0,-\deltay>[\nodea`\nodec;\labelb]%
\ignorespaces}%

\def\Dtrianglepairpppp(#1)|#2|/#3`#4`#5`#6`#7/<#8>[#9]{%
\def\next(##1,##2){\xpos##1\ypos##2}%
\next(#1)%
\def\next|##1##2##3##4##5|{\def\xa{##1}\def\xb{##2}%
\def\xc{##3}\def\xd{##4}\def\xe{##5}}%
\next|#2|%
\def\next<##1,##2>{\deltax=##1\deltay=##2\ignorespaces}%
\next<#8>%
\def\next[##1`##2`##3`##4;##5`##6`##7`##8`##9]{%
 \def\nodea{##1}\def\nodeb{##2}\def\nodec{##3}\def\noded{##4}%
 \def\labela{##5}\def\labelb{##6}\def\labelc{##7}\def\labeld{##8}\def\labele{##9}}%
\next[#9]%
\advance\ypos by \deltay%
\morphism(\xpos,\ypos)|\xc|/{#5}/<\deltax,0>[\nodeb`\nodec;\labelc]%
\morphism(\xpos,\ypos)|\xd|/{#6}/<0,-\deltay>[\nodeb`\noded;\labeld]%
\advance\ypos by \deltay%
\morphism(\xpos,\ypos)|\xa|/{#3}/<0,-\deltay>[\nodea`\nodeb;\labela]%
\morphism(\xpos,\ypos)|\xb|/{#4}/<\deltax,-\deltay>[\nodea`\nodec;\labelb]%
\advance\ypos by -\deltay%
\advance\xpos by \deltax%
\morphism(\xpos,\ypos)|\xe|/{#7}/<-\deltax,-\deltay>[\nodec`\noded;\labele]%
\ignorespaces}%

\def\Atrianglepair{\ifnextchar({\Atrianglepairp}{\Atrianglepairp(0,0)}}%
\def\Atrianglepairp(#1){\ifnextchar|{\Atrianglepairpp(#1)}%
{\Atrianglepairpp(#1)|lmrbb|}}%
\def\Atrianglepairpp(#1)|#2|{\ifnextchar/{\Atrianglepairppp(#1)|#2|}%
    {\Atrianglepairppp(#1)|#2|/>`>`>`>`>/}}%
\def\Atrianglepairppp(#1)|#2|/#3`#4`#5`#6`#7/{%
    \ifnextchar<{\Atrianglepairpppp(#1)|#2|/#3`#4`#5`#6`#7/}%
    {\Atrianglepairpppp(#1)|#2|/#3`#4`#5`#6`#7/<\default,\default>}}%

\def\Vtrianglepair{\ifnextchar({\Vtrianglepairp}{\Vtrianglepairp(0,0)}}%
\def\Vtrianglepairp(#1){\ifnextchar|{\Vtrianglepairpp(#1)}%
{\Vtrianglepairpp(#1)|aalmr|}}%
\def\Vtrianglepairpp(#1)|#2|{\ifnextchar/{\Vtrianglepairppp(#1)|#2|}%
    {\Vtrianglepairppp(#1)|#2|/>`>`>`>`>/}}%
\def\Vtrianglepairppp(#1)|#2|/#3`#4`#5`#6`#7/{%
    \ifnextchar<{\Vtrianglepairpppp(#1)|#2|/#3`#4`#5`#6`#7/}%
    {\Vtrianglepairpppp(#1)|#2|/#3`#4`#5`#6`#7/<\default,\default>}}%

\def\Ctrianglepair{\ifnextchar({\Ctrianglepairp}{\Ctrianglepairp(0,0)}}%
\def\Ctrianglepairp(#1){\ifnextchar|{\Ctrianglepairpp(#1)}%
{\Ctrianglepairpp(#1)|lrmlr|}}%
\def\Ctrianglepairpp(#1)|#2|{\ifnextchar/{\Ctrianglepairppp(#1)|#2|}%
    {\Ctrianglepairppp(#1)|#2|/>`>`>`>`>/}}%
\def\Ctrianglepairppp(#1)|#2|/#3`#4`#5`#6`#7/{%
    \ifnextchar<{\Ctrianglepairpppp(#1)|#2|/#3`#4`#5`#6`#7/}%
    {\Ctrianglepairpppp(#1)|#2|/#3`#4`#5`#6`#7/<\default,\default>}}%

\def\Dtrianglepair{\ifnextchar({\Dtrianglepairp}{\Dtrianglepairp(0,0)}}%
\def\Dtrianglepairp(#1){\ifnextchar|{\Dtrianglepairpp(#1)}%
{\Dtrianglepairpp(#1)|lrmlr|}}%
\def\Dtrianglepairpp(#1)|#2|{\ifnextchar/{\Dtrianglepairppp(#1)|#2|}%
    {\Dtrianglepairppp(#1)|#2|/>`>`>`>`>/}}%
\def\Dtrianglepairppp(#1)|#2|/#3`#4`#5`#6`#7/{%
    \ifnextchar<{\Dtrianglepairpppp(#1)|#2|/#3`#4`#5`#6`#7/}%
    {\Dtrianglepairpppp(#1)|#2|/#3`#4`#5`#6`#7/<\default,\default>}}%

\def\pplace[#1](#2,#3)[#4]{\POS(#2,#3)*+!!<0ex,\axis>!#1{#4}\ignorespaces}%
\def\cplace(#1,#2)[#3]{\POS(#1,#2)*+!!<0ex,\axis>{#3}\ignorespaces}%
\def\place{\ifnextchar[{\pplace}{\cplace}}

\def\pullback#1]#2]{\square#1]\trident#2]\ignorespaces}%

\def\tridentppp|#1#2#3|/#4`#5`#6/<#7,#8>[#9]{%
\def\next[##1;##2`##3`##4]{\def\nodee{##1}\def\labele{##2}%
   \def\labelf{##3}\def\labelg{##4}}%
\next[#9]%
\advance \xpos by -\deltax%
\advance \xpos by -#7\advance \ypos by #8%
\advance\deltax by #7%
\morphism(\xpos,\ypos)|#1|/{#4}/<\deltax,-#8>[\nodee`\nodeb;\labele]%
\advance\deltax by -#7%
\morphism(\xpos,\ypos)|#2|/{#5}/<#7,-#8>[\nodee`\nodea;\labelf]%
\advance\deltay by #8%
\morphism(\xpos,\ypos)|#3|/{#6}/<#7,-\deltay>[\nodee`\nodec;\labelg]%
\ignorespaces}%

\def\trident{\ifnextchar|{\tridentp}{\tridentp|amb|}}%
\def\tridentp|#1|{\ifnextchar/{\tridentpp|#1|}{\tridentpp|#1|/{>}`{>}`{>}/}}%
\def\tridentpp|#1|/#2/{\ifnextchar<{\tridentppp|#1|/#2/}%
  {\tridentppp|#1|/#2/<500,500>}}%

\def\setmorphismwidth#1#2#3#4{%
 \setbox0=\hbox{$#1{\labelstyle#3#3}#2$}#4=\wd0%
 \divide #4 by 2 \divide #4 by \ul%
 \advance #4 by 350 \ratchet{#4}{500}}%

\def\setSquarewidth[#1`#2`#3`#4;#5`#6`#7`#8]{%
 \setmorphismwidth{#1}{#2}{#5}{\topw}%
 \setmorphismwidth{#3}{#4}{#8}{\botw}%
\ratchet{\topw}{\botw}}%

\def\Squarepppp(#1)|#2|/#3/<#4>[#5]{%
 \setSquarewidth[#5]%
 \squarepppp(#1)|#2|/#3/<\topw,#4>[#5]%
\ignorespaces}%

\def\Square{\ifnextchar({\Squarep}{\Squarep(0,0)}}%
\def\Squarep(#1){\ifnextchar|{\Squarepp(#1)}{\Squarepp(#1)|alrb|}}%
\def\Squarepp(#1)|#2|{\ifnextchar/{\Squareppp(#1)|#2|}%
    {\Squareppp(#1)|#2|/>`>`>`>/}}%
\def\Squareppp(#1)|#2|/#3`#4`#5`#6/{%
    \ifnextchar<{\Squarepppp(#1)|#2|/#3`#4`#5`#6/}%
    {\Squarepppp(#1)|#2|/#3`#4`#5`#6/<\default>}}%

\def\hSquarespppp(#1,#2)|#3|/#4/<#5>[#6;#7]{%
\Xpos=#1\Ypos=#2%
\def\next|##1##2##3##4##5##6##7|{%
 \def\Xa{##1}\def\Xb{##2}\def\Xc{##3}\def\Xd{##4}%
 \def\Xe{##5}\def\Xf{##6}\def\Xg{##7}}%
\next|#3|%
\deltaY=#5%
\def\next[##1`##2`##3`##4`##5`##6]{%
 \def\Nodea{##1}\def\Nodeb{##2}\def\Nodec{##3}%
 \def\Noded{##4}\def\Nodee{##5}\def\Nodef{##6}}%
\next[#6]%
\def\next[##1`##2`##3`##4`##5`##6`##7]{%
 \def\Labela{##1}\def\Labelb{##2}\def\Labelc{##3}\def\Labeld{##4}%
 \def\Labele{##5}\def\Labelf{##6}\def\Labelg{##7}}%
\next[#7]%
\dohSquares/#4/}%

\def\dohSquares/#1`#2`#3`#4`#5`#6`#7/{%
\Squarepppp(\Xpos,\Ypos)|\Xa\Xc\Xd\Xf|/#1`#3`#4`#6/<\deltaY>%
 [\Nodea`\Nodeb`\Noded`\Nodee;\Labela`\Labelc`\Labeld`\Labelf]%
 \advance \Xpos by \topw
\Squarepppp(\Xpos,\Ypos)|\Xb\Xd\Xe\Xg|/#2``#5`#7/<\deltaY>%
[\Nodeb`\Nodec`\Nodee`\Nodef;\Labelb``\Labele`\Labelg]%
\ignorespaces}%

\def\hSquares{\ifnextchar({\hSquaresp}{\hSquaresp(0,0)}}%
\def\hSquaresp(#1){\ifnextchar|{\hSquarespp(#1)}{\hSquarespp%
(#1)|aalmrbb|}}%
\def\hSquarespp(#1)|#2|{\ifnextchar/{\hSquaresppp(#1)|#2|}%
    {\hSquaresppp(#1)|#2|/>`>`>`>`>`>`>/}}%
\def\hSquaresppp(#1)|#2|/#3/{%
    \ifnextchar<{\hSquarespppp(#1)|#2|/#3/}%
    {\hSquarespppp(#1)|#2|/#3/<\default>}}%

\def\vSquarespppp(#1,#2)|#3|/#4/<#5,#6>[#7;#8]{%
\Xpos=#1\Ypos=#2%
\def\next|##1##2##3##4##5##6##7|{%
 \def\Xa{##1}\def\Xb{##2}\def\Xc{##3}\def\Xd{##4}%
 \def\Xe{##5}\def\Xf{##6}\def\Xg{##7}}%
\next|#3|%
\deltaX=#5%
\deltaY=#6%
\def\next[##1`##2`##3`##4`##5`##6]{%
 \def\Nodea{##1}\def\Nodeb{##2}\def\Nodec{##3}%
 \def\Noded{##4}\def\Nodee{##5}\def\Nodef{##6}}%
\next[#7]%
\def\next[##1`##2`##3`##4`##5`##6`##7]{%
 \def\Labela{##1}\def\Labelb{##2}\def\Labelc{##3}\def\Labeld{##4}%
 \def\Labele{##5}\def\Labelf{##6}\def\Labelg{##7}}%
\next[#8]%
\dovSquares/#4/\ignorespaces}%

\def\dovSquares/#1`#2`#3`#4`#5`#6`#7/{%
\setmorphismwidth{\Nodea}{\Nodeb}{\Labela}{\topw}%
\setmorphismwidth{\Nodec}{\Noded}{\Labeld}{\botw}%
\ratchet{\topw}{\botw}%
\setmorphismwidth{\Nodee}{\Nodef}{\Labelg}{\botw}%
\ratchet{\topw}{\botw}%
\square(\Xpos,\Ypos)|\Xd\Xe\Xf\Xg|/`#5`#6`#7/<\topw,\deltaX>%
 [\Nodec`\Noded`\Nodee`\Nodef;`\Labele`\Labelf`\Labelg]%
\advance \Ypos by \deltaX%
\square(\Xpos,\Ypos)|\Xa\Xb\Xc\Xd|/#1`#2`#3`#4/<\topw,\deltaY>%
 [\Nodea`\Nodeb`\Nodec`\Noded;\Labela`\Labelb`\Labelc`\Labeld]%
}%

\def\vSquares{\ifnextchar({\vSquaresp}{\vSquaresp(0,0)}}%
\def\vSquaresp(#1){\ifnextchar|{\vSquarespp(#1)}{\vSquarespp%
(#1)|alrmlrb|}}%
\def\vSquarespp(#1)|#2|{\ifnextchar/{\vSquaresppp(#1)|#2|}%
    {\vSquaresppp(#1)|#2|/>`>`>`>`>`>`>/}}%
\def\vSquaresppp(#1)|#2|/#3/{%
    \ifnextchar<{\vSquarespppp(#1)|#2|/#3/}%
    {\vSquarespppp(#1)|#2|/#3/<\default,\default>}}%

\def\osquarepppp(#1)|#2|/#3`#4`#5`#6/<#7>[#8]{\squarepppp%
 (#1)|#2|/#3`#4`#5`#6/<#7>[#8]%
 \let\Nodea\nodea\let\Nodeb\nodeb%
\let\Nodec\nodec\let\Noded\noded\Xpos=\xpos\Ypos=\ypos%
\deltaX=\deltax \deltaY=\deltay \isquare}

\def\cube{\ifnextchar({\osquarep}{\osquarep(0,0)}}%
\def\osquarep(#1){\ifnextchar|{\osquarepp(#1)}{\osquarepp(#1)|alrb|}}%
\def\osquarepp(#1)|#2|{\ifnextchar/{\osquareppp(#1)|#2|}%
    {\osquareppp(#1)|#2|/>`>`>`>/}}%
\def\osquareppp(#1)|#2|/#3`#4`#5`#6/{%
    \ifnextchar<{\osquarepppp(#1)|#2|/#3`#4`#5`#6/}%
    {\osquarepppp(#1)|#2|/#3`#4`#5`#6/<1500,1500>}}%

\def\isquarepppp(#1)|#2|/#3`#4`#5`#6/<#7>[#8]{%
 \squarepppp(#1)|#2|/#3`#4`#5`#6/<#7>[#8]%
\ifnextchar|{\cubep}{\cubep|mmmm|}}%
\def\cubep|#1|{\ifnextchar/{\cubepp|#1|}{\cubepp|#1|/>`>`>`>/}}%

\def\isquare{\ifnextchar({\isquarep}{\isquarep(\default,\default)}}%
\def\isquarep(#1){\ifnextchar|{\isquarepp(#1)}{\isquarepp(#1)|alrb|}}
\def\isquarepp(#1)|#2|{\ifnextchar/{\isquareppp(#1)|#2|}%
    {\isquareppp(#1)|#2|/>`>`>`>/}}%
\def\isquareppp(#1)|#2|/#3`#4`#5`#6/{%
    \ifnextchar<{\isquarepppp(#1)|#2|/#3`#4`#5`#6/}%
    {\isquarepppp(#1)|#2|/#3`#4`#5`#6/<500,500>}}%

\def\cubepp|#1#2#3#4|/#5`#6`#7`#8/[#9]{%
\def\next[##1`##2`##3`##4]{\gdef\Labela{##1}%
\gdef\Labelb{##2}\gdef\Labelc{##3}\gdef\Labeld{##4}}\next[#9]%
\xend\xpos \yend\ypos
\Xend\xend\advance\Xend by -\Xpos
\Yend\yend\advance\Yend by -\Ypos
\domorphism(\Xpos,\Ypos)|#2|/#6/<\Xend,\Yend>[\Nodeb`\nodeb;\Labelb]%
\advance\Xpos by-\deltaX
\advance\xend by-\deltax
\Xend\xend\advance\Xend by -\Xpos
\domorphism(\Xpos,\Ypos)|#1|/#5/<\Xend,\Yend>[\Nodea`\nodea;\Labela]%
\advance\Ypos by-\deltaY
\advance\yend by-\deltay
\Yend\yend\advance\Yend by -\Ypos
\domorphism(\Xpos,\Ypos)|#3|/#7/<\Xend,\Yend>[\Nodec`\nodec;\Labelc]%
\advance\Xpos by\deltaX
\advance\xend by\deltax
\Xend\xend\advance\Xend by -\Xpos
\domorphism(\Xpos,\Ypos)|#4|/#8/<\Xend,\Yend>[\Noded`\noded;\Labeld]%
\ignorespaces}

\def\setwdth#1#2{\setbox0\hbox{$\labelstyle#1$}\wdth=\wd0
\setbox0\hbox{$\labelstyle#2$}\ifnum\wdth<\wd0 \wdth=\wd0 \fi}

\def\topppp/#1/<#2>^#3_#4{\allowbreak\mathrel{%
\ifnum#2=0
   \setwdth{#3}{#4}\deltax=\wdth \divide \deltax by \ul
   \advance \deltax by \defaultmargin  \ratchet{\deltax}{200}%
\else \deltax #2
\fi
\xy\ar@{#1}^{#3}_{#4}(\deltax,0) \endxy
\ignorespaces}}

\def\toppp/#1/<#2>^#3{\ifnextchar_{\topppp/#1/<#2>^{#3}}{\topppp/#1/<#2>^{#3}_{}}}
\def\topp/#1/<#2>{\ifnextchar^{\toppp/#1/<#2>}{\toppp/#1/<#2>^{}}}
\def\toop/#1/{\ifnextchar<{\topp/#1/}{\topp/#1/<0>}}
\def\to{\ifnextchar/{\toop}{\toop/>/}}


\def\mon{\to/ >->/}
\def\epi{\to/->>/}
\def\toleft{\to/<-/}
\def\monleft{\to/<-< /}
\def\epileft{\to/<<-/}


\def\twopppp/#1`#2/<#3>^#4_#5{\allowbreak\mathrel{%
\ifnum0=#3
  \setwdth{#4}{#5}\deltax=\wdth \divide \deltax by \ul \advance \deltax
  by \defaultmargin \ratchet{\deltax}{200}%
\else \deltax#3 \fi
\xy\ar@{#1}@<2.5pt>^{#4}(\deltax,0)%
\ar@{#2}@<-2.5pt>_{#5}(\deltax,0)\endxy\ignorespaces}}

\def\twoppp/#1`#2/<#3>^#4{\ifnextchar_{\twopppp/#1`#2/<#3>^{#4}}%
  {\twopppp/#1`#2/<#3>^{#4}_{}}}
\def\twopp/#1`#2/<#3>{\ifnextchar^{\twoppp/#1`#2/<#3>}{\twoppp/#1`#2/<#3>^{}}}
\def\twop/#1`#2/{\ifnextchar<{\twopp/#1`#2/}{\twopp/#1`#2/<0>}}
\def\two{\ifnextchar/{\twop}{\twop/>`>/}}

\def\twoleft{\two/<-`<-/}

\def\threeppppp/#1`#2`#3/<#4>^#5|#6_#7{\allowbreak\mathrel{%
\ifnum0=#4
\setbox0\hbox{$\labelstyle#5$}\wdth=\wd0
\setbox0\hbox{$\labelstyle#6$}\ifnum\wdth<\wd0 \wdth=\wd0 \fi
\setbox0\hbox{$\labelstyle#7$}\ifnum\wdth<\wd0 \wdth=\wd0 \fi
\deltax=\wdth \divide \deltax by \ul \advance \deltax by
\defaultmargin \ratchet{\deltax}{300}%
\else\deltax#4 \fi
    \xy \ifnum\wd0=0 \ar@{#2}(\deltax,0)
    \else \ar@{#2}|{#6}(\deltax,0)\fi
\ar@{#1}@<4.5pt>^{#5}(\deltax,0)
\ar@{#3}@<-4.5pt>_{#7}(\deltax,0)\endxy\ignorespaces}}

\def\threepppp/#1`#2`#3/<#4>^#5|#6{\ifnextchar_{\threeppppp
  /#1`#2`#3/<#4>^{#5}|{#6}}{\threeppppp/#1`#2`#3/<#4>^{#5}|{#6}_{}}}
\def\threeppp/#1`#2`#3/<#4>^#5{\ifnextchar|{\threepppp
  /#1`#2`#3/<#4>^{#5}}{\threepppp/#1`#2`#3/<#4>^{#5}|{}}}
\def\threepp/#1`#2`#3/<#4>{\ifnextchar^{\threeppp/#1`#2`#3/<#4>}%
  {\threeppp/#1`#2`#3/<#4>^{}}}
\def\threep/#1`#2`#3/{\ifnextchar<{\threepp/#1`#2`#3/}%
  {\threepp/#1`#2`#3/<0>}}
\def\three{\ifnextchar/{\threep}{\threep/>`>`>/}}


\def\twoar(#1,#2){{%
 \scalefactor{0.1}
 \deltax#1\deltay#2%
 \deltaX=\ifnum\deltax<0-\fi\deltax
 \deltaY=\ifnum\deltay<0-\fi\deltay
 \Xend\deltax \multiply \Xend by \deltax
 \Yend\deltay \multiply \Yend by \deltay
 \advance\Xend by \Yend \multiply \Xend by 3
 \ifnum \deltaX > \deltaY
    \multiply \deltaX by 3 \advance \deltaX by \deltaY
 \else
    \multiply \deltaY by 3 \advance \deltaX by \deltaY
 \fi
 \multiply\deltax by 500
 \multiply\deltay by 500
 \xpos\deltax \multiply \xpos by 3 \divide\xpos by \deltaX
 \Xpos\deltax \multiply \Xpos by \deltaX \divide \Xpos by \Xend
 \advance \xpos by \Xpos
 \ypos\deltay \multiply \ypos by 3 \divide\ypos by \deltaX
 \Ypos\deltay \multiply \Ypos by \deltaX \divide \Ypos by \Xend
 \advance \ypos by \Ypos
 \xy \ar@{=>}(\xpos,\ypos) \endxy
}\ignorespaces}

\def\iiixiiipppppp(#1,#2)|#3|/#4/<#5>#6<#7>[#8;#9]{%
 \xpos#1\ypos#2\relax
 \def\next|##1##2##3##4##5##6##7|{\def\xa{##1}\def\xb{##2}%
 \def\xc{##3}\def\xd{##4}\def\xe{##5}\def\xf{##6}\nextt|##7|}%
 \def\nextt|##1##2##3##4##5##6|{\def\xg{##1}\def\xh{##2}%
 \def\xi{##3}\def\xj{##4}\def\xk{##5}\def\xl{##6}}%
 \next|#3|%
 \def\next<##1,##2>{\deltax##1\deltay##2}%
 \next<#5>%
 \def\next<##1,##2>{\deltaX##1\deltaY##2}%
 \next<#7>%
 \def\next##1{\topw##1\relax
 \ifodd\topw \def\zl{}\else\def\zl{\relax}\fi \divide\topw by 2
 \ifodd\topw \def\zk{}\else\def\zk{\relax}\fi \divide\topw by 2
 \ifodd\topw \def\zj{}\else\def\zj{\relax}\fi \divide\topw by 2
 \ifodd\topw \def\zi{}\else\def\zi{\relax}\fi \divide\topw by 2
 \ifodd\topw \def\zh{}\else\def\zh{\relax}\fi \divide\topw by 2
 \ifodd\topw \def\zg{}\else\def\zg{\relax}\fi \divide\topw by 2
 \ifodd\topw \def\zf{}\else\def\zf{\relax}\fi \divide\topw by 2
 \ifodd\topw \def\ze{}\else\def\ze{\relax}\fi \divide\topw by 2
 \ifodd\topw \def\zd{}\else\def\zd{\relax}\fi \divide\topw by 2
 \ifodd\topw \def\zc{}\else\def\zc{\relax}\fi \divide\topw by 2
 \ifodd\topw \def\zb{}\else\def\zb{\relax}\fi \divide\topw by 2
 \ifodd\topw \def\za{}\else\def\za{\relax}\fi}%
 \next{#6}%
 \def\next[##1`##2`##3`##4`##5`##6`##7`##8`##9]{%
 \def\nodea{##1}\def\nodeb{##2}\def\nodec{##3}%
 \def\noded{##4}\def\nodee{##5}\def\nodef{##6}%
 \def\nodeg{##7}\def\nodeh{##8}\def\nodei{##9}}%
 \next[#8]%
 \def\next[##1`##2`##3`##4`##5`##6`##7]{%
 \def\labela{##1}\def\labelb{##2}\def\labelc{##3}%
 \def\labeld{##4}\def\labele{##5}\def\labelf{##6}\nextt[##7]}%
 \def\nextt[##1`##2`##3`##4`##5`##6]{%
 \def\labelg{##1}\def\labelh{##2}\def\labeli{##3}%
 \def\labelj{##4}\def\labelk{##5}\def\labell{##6}}%
 \next[#9]%
 \def\next/##1`##2`##3`##4`##5`##6`##7/{%
\morphism(\xpos,\ypos)|\xe|/{##5}/<\deltax,0>[\nodeg`\nodeh;\labele]%
 \ifx\zi\empty\relax \morphism(\xpos,\ypos)||/<-/<-\deltaX,0>[\nodeg`0;]\fi
 \ifx\zd\empty\relax \morphism(\xpos,\ypos)||<0,-\deltaY>[\nodeg`0;]\fi
 \advance\xpos by \deltax
 \morphism(\xpos,\ypos)|\xf|/{##6}/<\deltax,0>[\nodeh`\nodei;\labelf]%
 \ifx\ze\empty\relax \morphism(\xpos,\ypos)||<0,-\deltaY>[\nodeh`0;]\fi
 \advance\xpos by \deltax
 \ifx\zf\empty\relax \morphism(\xpos,\ypos)||<0,-\deltaY>[\nodei`0;]\fi
 \ifx\zl\empty\relax \morphism(\xpos,\ypos)||<\deltaX,0>[\nodei`0;]\fi
 \advance\ypos by \deltay
 \ifx\zk\empty\relax \morphism(\xpos,\ypos)||<\deltaX,0>[\nodef`0;]\fi
 \advance\xpos by -\deltax
 \morphism(\xpos,\ypos)|\xd|/{##4}/<\deltax,0>[\nodee`\nodef;\labeld]%
 \advance\xpos by -\deltax
 \morphism(\xpos,\ypos)|\xc|/{##3}/<\deltax,0>[\noded`\nodee;\labelc]%
 \ifx\zh\empty\relax \morphism(\xpos,\ypos)||/<-/<-\deltaX,0>[\noded`0;]\fi
 \advance\ypos by \deltay
 \morphism(\xpos,\ypos)|\xa|/{##1}/<\deltax,0>[\nodea`\nodeb;\labela]%
 \ifx\zg\empty\relax \morphism(\xpos,\ypos)||/<-/<-\deltaX,0>[\nodea`0;]\fi
 \ifx\za\empty\relax \morphism(\xpos,\ypos)||/<-/<0,\deltaY>[\nodea`0;]\fi
 \advance\xpos by \deltax
 \morphism(\xpos,\ypos)|\xb|/{##2}/<\deltax,0>[\nodeb`\nodec;\labelb]%
 \ifx\zb\empty\relax \morphism(\xpos,\ypos)||/<-/<0,\deltaY>[\nodeb`0;]\fi
 \advance\xpos by \deltax
 \ifx\zc\empty\relax \morphism(\xpos,\ypos)||/<-/<0,\deltaY>[\nodec`0;]\fi
 \ifx\zj\empty\relax \morphism(\xpos,\ypos)||<\deltaX,0>[\nodec`0;]\fi
 \nextt/##7/}%
 \def\nextt/##1`##2`##3`##4`##5`##6/{%
 \morphism(\xpos,\ypos)|\xi|/{##3}/<0,-\deltay>[\nodec`\nodef;\labeli]%
 \advance\xpos by -\deltax
 \morphism(\xpos,\ypos)|\xh|/{##2}/<0,-\deltay>[\nodeb`\nodee;\labelh]%
 \advance\xpos by -\deltax
 \morphism(\xpos,\ypos)|\xg|/{##1}/<0,-\deltay>[\nodea`\noded;\labelg]%
 \advance\ypos by -\deltay
 \morphism(\xpos,\ypos)|\xj|/{##4}/<0,-\deltay>[\noded`\nodeg;\labelj]%
 \advance\xpos by \deltax
 \morphism(\xpos,\ypos)|\xk|/{##5}/<0,-\deltay>[\nodee`\nodeh;\labelk]%
 \advance\xpos by \deltax
 \morphism(\xpos,\ypos)|\xl|/{##6}/<0,-\deltay>[\nodef`\nodei;\labell]}%
 \next/#4/\ignorespaces}

\def\iiixiii{\ifnextchar({\iiixiiip}{\iiixiiip(0,0)}}%
\def\iiixiiip(#1){\ifnextchar|{\iiixiiipp(#1)}%
  {\iiixiiipp(#1)|aammbblmrlmr|}}%
\def\iiixiiipp(#1)|#2|{\ifnextchar/{\iiixiiippp(#1)|#2|}%
    {\iiixiiippp(#1)|#2|/>`>`>`>`>`>`>`>`>`>`>`>/}}%
\def\iiixiiippp(#1)|#2|/#3/{%
    \ifnextchar<{\iiixiiipppp(#1)|#2|/#3/}%
    {\iiixiiipppp(#1)|#2|/#3/<\default,\default>}}%
\def\iiixiiipppp(#1)|#2|/#3/<#4>{\ifnextchar[{\iiixiiippppp(#1)|#2|/#3/%
   <#4>0<0,0>}{\iiixiiippppp(#1)|#2|/#3/<#4>}}%
\def\iiixiiippppp(#1)|#2|/#3/<#4>#5{\ifnextchar<%
   {\iiixiiipppppp(#1)|#2|/#3/<#4>{#5}}%
   {\iiixiiipppppp(#1)|#2|/#3/<#4>{#5}<400,400>}}%

\def\iiixiipppppp(#1,#2)|#3|/#4/<#5>#6<#7>[#8;#9]{%
 \xpos#1\ypos#2\relax
 \def\next|##1##2##3##4##5##6##7|{\def\xa{##1}\def\xb{##2}%
 \def\xc{##3}\def\xd{##4}\def\xe{##5}\def\xf{##6}\def\xg{##7}}%
 \next|#3|%
 \def\next<##1,##2>{\deltax##1\deltay##2}%
 \next<#5>%
 \deltaX#7
 \topw#6
 \def\next{%
 \ifodd\topw \def\za{}\else\def\za{\relax}\fi \divide\topw by 2
 \ifodd\topw \def\zb{}\else\def\zb{\relax}\fi \divide\topw by 2
 \ifodd\topw \def\zc{}\else\def\zc{\relax}\fi \divide\topw by 2
 \ifodd\topw \def\zd{}\else\def\zd{\relax}\fi}%
 \next
 \def\next[##1`##2`##3`##4`##5`##6]{%
 \def\nodea{##1}\def\nodeb{##2}\def\nodec{##3}%
 \def\noded{##4}\def\nodee{##5}\def\nodef{##6}}%
 \next[#8]%
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SLTPinQL3.bbl
\begin{thebibliography}{10}

\bibitem{Aerts:1982}
D.~Aerts.
\newblock Description of many separated physical entities without the paradoxes
  encountered in quantum mechanics.
\newblock {\em Found. Phys.}, 12(12):1131--1170, 1982.

\bibitem{AertsHPA84}
D.~Aerts.
\newblock The missing elements of reality in the description of quantum
  mechanics of the {EPR} paradox situation.
\newblock {\em Helv. Phys. Acta}, 57(4):421--428, 1984.

\bibitem{Barr:handbook}
M.~Barr.
\newblock {\em {$\ast$}-autonomous categories}, volume 752 of {\em Lecture
  Notes in Mathematics}.
\newblock Springer, Berlin, 1979.
\newblock With an appendix by Po Hsiang Chu.

\bibitem{Birkhoff/Neumann:1936}
G.~Birkhoff and J.~v. Neumann.
\newblock The logic of quantum mechanics.
\newblock {\em Ann. of Math. (2)}, 37(4):823--843, 1936.

\bibitem{Faure/Froelicher:handbook}
C.-A. Faure and A.~Fr{\"o}licher.
\newblock {\em Modern projective geometry}, volume 521 of {\em Mathematics and
  its Applications}.
\newblock Kluwer Academic Publishers, Dordrecht, 2000.

\bibitem{Faure/Moore/Piron:1995}
C.-A. Faure, D.~J. Moore, and C.~Piron.
\newblock Deterministic evolutions and {S}chr\"odinger flows.
\newblock {\em Helv. Phys. Acta}, 68(2):150--157, 1995.

\bibitem{Fraser:1976}
G.~A. Fraser.
\newblock The semilattice tensor product of distributive lattices.
\newblock {\em Trans. Amer. Math. Soc.}, 217:183--194, 1976.

\bibitem{Golfin:handbook}
A.~S. Golfin.
\newblock {\em Representations and products of lattices}.
\newblock PhD thesis, U. Mass., Amherts, 1987.

\bibitem{Graetzer/Wehrung:1999}
G.~Gr{\"a}tzer and F.~Wehrung.
\newblock A new lattice construction: the box product.
\newblock {\em J. Algebra}, 221(1):315--344, 1999.

\bibitem{Ischi(rmp):2004}
B.~Ischi.
\newblock A characterization of the {A}erts product of {H}ilbertian lattices.
\newblock Submitted, available on \verb"http://arxiv.org/abs/math-ph/0405048".

\bibitem{IschiDenver:2004}
B.~Ischi.
\newblock Orthocomplementation and compound systems.
\newblock Submitted, available on \verb"http://"
  \verb"arxiv.org/abs/quant-ph/0410085".

\bibitem{Ischi:2000}
B.~Ischi.
\newblock Endomorphisms of the separated product of lattices.
\newblock {\em Internat. J. Theoret. Phys.}, 39(11):2559--2581, 2000.

\bibitem{Maeda/Maeda:handbook}
F.~Maeda and S.~Maeda.
\newblock {\em Theory of symmetric lattices}.
\newblock Die Grundlehren der mathematischen Wissenschaften, Band 173.
  Springer-Verlag, New York, 1970.

\bibitem{vonNeumann:handbook}
J.~v. Neumann.
\newblock {\em Mathematical foundations of quantum mechanics}.
\newblock Princeton University Press, Princeton, 1955.
\newblock Translated by Robert T. Beyer.

\bibitem{Piron:1964}
C.~Piron.
\newblock Axiomatique quantique.
\newblock {\em Helv. Phys. Acta}, 37:439--468, 1964.

\bibitem{Piron:handbook}
C.~Piron.
\newblock {\em Foundations of quantum physics}.
\newblock W. A. Benjamin, Inc., Advanced Book Program, Reading,
  Mass.-London-Amsterdam, 1976.
\newblock Mathematical Physics Monograph Series, 19.

\bibitem{Pulmannova:1985}
S.~Pulmannov{\'a}.
\newblock Tensor product of quantum logics.
\newblock {\em J. Math. Phys.}, 26(1):1--5, 1985.

\bibitem{Seal:2004}
G.~J. Seal.
\newblock Cartesian closed topological categories and tensor products.
\newblock {\em Appl. Categ. Structures}, 13(1):37--37, 2005.

\bibitem{Shmuely:1974}
Z.~Shmuely.
\newblock The structure of {G}alois connections.
\newblock {\em Pacific J. Math.}, 54(2):209--225, 1974.

\end{thebibliography}
