Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $L$ be an atomic ortholattice. We say that two elements $a$ and $b$ of $L$ are orthogonal if $a\leq b^\perp$. If $L$ is orthomodular then every element of $L$ can be written as a join of pairwise orthogonal atoms of $L$. Is the converse also true?

share|improve this question

2 Answers 2

up vote 2 down vote accepted

Yes there are. Any non-modular finite orthomodular lattice is an example.

A concrete example: take the Boolean algebras $2^2$,$2^3$ and identify their top and bottom elements. This is clearly an atomistic ortholattice, which is not modular, because it contains a pentagon as a sublattice.

share|improve this answer

Examples are contained in a classical book about semimodular lattice theory: F. Maeda and S. Maeda "theory of symmetric lattices". See that book for details and refernces concerning what follows.

Let $V$ be a infinite dimensional, separable, metrically incomplete pre-Hilbert space (over real, complex or quaternionic scalars, or more generally any archimedean Baer-ordered $*$-field). Let $L$ the collection of orthoclosed subspaces of $V$ (orthocosed means being the orthogonal of a subset of $V$, i.e. being the biorthogonal of itself).

Ordered by inclusion, $L$ is a complete lattice (the lattice of ``closed'' set for the Galois connection given by orthogonality). $L$ is orthocomplemented in the obvious way (the orthocomplement is the set of orthogonal vectors). $L$ is atomistic (every element is join of atoms), with atoms being the one dimensional linear subspaces of $V$ (the finite dimensional linear subspaces are all orthoclosed, form a join dense modular sublattice, and together with their orthocomplements form a modular orthocomplemented lattice whose completion by cuts is canonically
identified with $L$).

The babylonian method for completing the square (m\'ethode connue \'egalment, dans dans la litt\'erature, sous le noms de "m\'ethode de Jacobi" et "m\'ethode d'ortogonalization de Schmidt", `a moins qu'il ne conviene plut\^ot de l'attribuer `a quelque savant russe, wrote A. Weil in 1957) shows that each separable (and also each complete)
pre-Hilbert space has an orthogonal basis (but not always so in every non-separable pre-Hilbert space). Hence every element in $L$ is a orthogonal join of atoms.

To show that $L$ is not orthomodular precisely because it is metrically incomplete, now apply the theorem of Piron - Amemya / Araki - Morash - Gross / Keller - Wilbur - Holland.

(Digression: One does not need the much deeper and mathematically wonderful theorem of M. P. Soler; neither one does need the even deeper, and unfortunately well forgotten by modern quantum logicians, characterization by von Neumann of "continuous geometries with transition probability". This characterization, even when applied to its easier subcase, the type I finite case, provides an axiomatization of Hilberian quantum logics of lenght at lest 4 which
physically much more meaningful that the one obtained by modern quantum logicians form Soler's theorem, which is moreover restricted to the infinite dimensional case only. End digression).

The above example $L$ has infinite lengh. I have no well known examples of finite lenght (I am familiar only with the semimodular case, which implies modularity in finite lenght with anti-automorphism). However, some related comments follow, ending in a candidate example.

Note that the usual meaning of "atomistic" is: each element is join of atoms. Equivalently (see again Maeda - Maeda): atomic (each nonzero element contains an atom) and sectionally semi-complemented (if $x<y$ then $y$ contains a nonzero $z$ disjoint from $x$).

Since "orthomodular" means "relatively orthocomplemented" (if $x<y$ then a $z$ orthogonal to $x$ exists whose join with $x$ is $y$), one sees that "atomic orthomodular" imples atomistic. Besides, for complete orthocomplemented atomic lattices, one sees that orthomodular is equivalent to "each set of orthogonal atoms contained in a element $x$ expands to one such set with join $x$", and is also equivalent to "each maximal orthogonal set of atoms contained in a element $x$ has join $x$". These equivalent conditions for orthomodularity are strictly stronger than "each element is a orthogonal join of atoms" as shown by incomplete separable pre-Hilbert spaces.

On the other hand, atomistic is strictly weaker than "each element is a join of orthogonal atoms". Infact, the following method produces finite, atomistic orthocomplemented lattices which are not orthomodular because some elements have no orthogonal basis of atoms.

Take any finite orthoalgebra $A$ which is not a lattice (examples are frequently given in literature, using Greechie diagrams). Let $L$ be its cut (normal, Dedekind - McNeille) completion. $L$ is complete, even finite, othocomplemented lattice (as any completion of a orthocomplemented, even finite, orthoposet $A$). $L$ is atomistic since
$A$ is atomistic and join dense in $L$; this also implies that $A$ and
$L$ have the same atoms. Hence also the orthogonal joins of atoms are the same in $A$ and in $L$; such joins are exactly the elements of $A$. So, each element of $L$ not in $A$ does not have a orthogonal basis of atoms.

Two final, and formally off-topic (but hopefully intuitively helpful) remarks (before the appendix with the candidate example).

First, one can easy also give atomistic orthocomplemented examples where there are many elements besides atoms and the top / bottom
elements, but all these elements have no orthogonal basis of atoms (even no orthogonal splitting in nonzero elements). Take any atomistic and coatomistic lattice $A$, and define $L$ as the 0-1 pasting (horizontal sum) of $A$ with its dual $A'$ (the bottom element of $A$ and $A'$ are identified; the same for the top, but otherwise the set-union is disjoint and so is the order relation; in particoular, a pair of non-extreme elements one in $A$ and the orther in $A'$ is a complementary pair). $L$ is orthocomplemented by taking as orthocompelementary pairs the $a,a'$ with $a$ in $A$ and as $a'$ the corresponding element as seen in the dual $A'$. So any two orthogonal atoms of $L$ have the top element as join.

Second remark. To any orthocomplemented lattice $L$ one can associate at least two orthomodular posets (which coincide with $L$ iff $L$ is orthomodular). First, consider the same set $L$, but this time with a stronger ordering: $x$ has a orthocomplement in $y$ (as computed in $L$; one could say: $x$ splits $y$). One obtains a $L'$ contained'' in $L$ (same elements and ortogonality, but a smaller ordering relation). One obtains a even smaller $L''$ using only thesplitting'' elements of $L$ (the elements that are orthocomplemented in each larger element of $L$); it is a sub-orthoposet of $L$ and a sub-orthomodular-poset of $L'$. In the usual atomic cases (associated to linear orthogonality
relations in projective geometries) one has that finite joins of atoms form a join-dense lattice ideal of $L$ contained in $L''$; as shown above, it is not always so in general.

If there is a finite lenght orthocomplementd lattice $L$, not orthomodular but with orthogonal bases of atoms for each element, then comparing $L$ with $L'$ and $L''$ should be instructive, and so should be comparing with the set of elements which have a orthogonal basis of atoms (note that atoms give a Foulis - Randall manual, with a associated orthoalgebra consisting exactly of such elements).

[Form Henstock (his general theory of integration, in division spaces), a
illuminating non-example: in the cartesian plane, consider figures that "modulo the border" are finite disjoint unions of rectangles with sides either parallel to the axes or to the bisectors of the axes (formally, work in the boolean algebra of regular open sets, or equivalently open / closed sets modulo nowhere dense sets). It is a non-example since (besides being atomless) it has no top element and no joins despite being sup-directed; it is instructive since one has two unitless Boolean rings "accidentally pasted together" using the inclusion order in the plane, a order which has nothing to say about the "intrinsic" state-induced order (the associated $L'$, a poset whose bounded intervals are Boolean algebras). A $L$, if it exists, would be in same way analogous: the orthomodular structures associated to $L$ might be the horizontal sum of two finite boolean algebras, but $L$ itself has more pairs in the ordering relation, with "non splitting
atoms" inside an element that kill orthomodularity of $L$.]

Appendix:

Tentative candidate finite counterexample: first, take the horizontal sum of two copies of the boolean algebra with two atoms $2^3$ (or more generally two finite orthomodular lattices $A$, with an atom $a$ whose orthocomplement $a'$ is not an atom, and $B$, with an atom $b$ whose orthocomplement $b'$ is not an atom). Then, to this orthomodular
horizontal sum add two pairs to the ordering relation: $a<b'$ and $b<a'$. If this is still a lattice (isn't it? very few joins and meets should change), then it should be orthocomplementd (i.e. meet and join
of pairs $x,x'$ do not change), not orthomodular ($a$ does not split $b'$) but each element has a orthogonal basis of atoms (any basis in the horizontal sum).

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.