ORTHOMODULAR LATTICES
By
RICHARD JOSEPH GREECHIE
A DISSERTATION PRESENTED TO THE GRADUATE COUNCIL OF
THE UNIVERSITY OF FLORIDA
IN PARTIAL FULFILLMENT OF THE REQUIREMENTS FOR THE
DEGREE OF DOCTOR OF PHILOSOPHY
UNIVERSITY OF FLORIDA
APRIL, 1966
ACKNOWLEDGEMENT
The author wishes to acknowledge the helpful
suggestions, encouragement, and superior example of his
advisor, Professor David Foulis.
TABLE OF CONTENTS
page
ACKNOWLEDGEMENT . . . . .. . . . ii
LIST OF FIGURES . . . .. . . . . . iv
INDEX OF SYMBOLS . . . .. . . . . v
INDEX OF TERMINOLOGY . . . . . . ... . vi
CHAPTER
1. DEFINITIONS AND ELEMENTARY RESULTS . . .. 1
1. From Relations to Orthomodular Lattices 1
2. Standard Results in the Theory
of Orthomodular Lattices . . . . 9
3. Morphisms, Ideals, Filters, and Sections 16
2. THE PASTE JOB . . . . . . . .. 21
1. New Lattices from Old . . . . .. 21
2. A Partial Converse . . . . .. 47
3. Variation on a Theme . . . . .. 54
3. THE REPRESENTATION OF ORTHOCOMPLEMENTED
POSETS BY SETS . . . . . . ... 70
1. Fsequences . . . . . . ... 70
2. Order Orthohomomorphisms . . . .. 79
3. Travis Matrices . . . . . .. 83
4. Examples . . . . . . . .. 87
BIBLIOGRAPHY . . . . . . . ... . 93
BIOGRAPHICAL SKETCH . . . . . . . .. 94
LIST OF FIGURES
page
Figure 1 . . . . . . . . 10
Figure 2 . . . . . . . ... 40
Figure 3 . . . . . . . ... 42
Figure 4 . . . . . . ... 42
Figure 5 . . . . . . . ... 43
Figure 6 . . . . . . . ... 45
Figure 7 . . . . . . . ... 46
Figure 8 . . . . . . . ... 65
Figure 9 . . . . . . . ... 65
Figure 10 . . . . . . . ... 66
Figure 11 . . . . . . . ... 66
Figure 12 . . . . . . ... 87
Figure 13 . . . . . . . ... 88
Figure 14 . . . . . . . ... 88
Figure 15 . . . . . . . ... 89
Figure 16 . . . . . . . ... 90
Figure 17 . . . . . . . ... 91
Figure 18 . . . . . . ... 92
INDEX OF SYMBOLS
page
M . . 1
2, 2 . 1
E . . . 1
C, 3 . .. 1
MN . . 1
T.A.E. . 1
XXX . . 1
R1 . . .
Aly . . 2
2 3
R2, R . 2
X/R . . 3
<, < . . 3
(X,R) . . 3
U(M), T(M) 4
sup M, inf M 4
, A . . 5
0, 1 . . 6
P(0,x), P(x,1
':P > P
(P,<,')
J_ . . .
OMI . .
DOMI . .
Cp . . ..
C . . .
C(M) . .
Pol(M) . .
D(a,b,c) .
M(b,c) . .
S . . .
x
[x] . .
[z]:[x]R[y]
[z]:[x] r [y]
page
). 6
7
S 7
S 9
. 9
S. 10
S. 11
S. 11
S. 11
S. 13
S. 15
S. 15
S. 17
S. 19
S. 22
S. 22
S 24
[0,1)i .
HS . .
D 16 .
N, R
[0,1]
1 2
fM . .
x
B . .
B
Y
Y
bi, B
[y] . .
. .
H, E, F
(2*,p*, ')
5* . .
page
30
* 54
66
71
S71
71
. 71
S71
. 72
. 72
S72
. 74
S74
. 74
80
S83
S83
INDEX OF TERMINOLOGY
age
adjunction of a crown .
antisymmetric . . .
atom . . . . .
atomic . . . . .
Boolean . . . . .
center . . . . .
chain . . . . .
coatom . . . . .
commutes . . . .
comparable . . . .
complement . . . .
complemented poset . .
complete join
homomorphism . . .
complete joinmeet
homomorphism . . .
complete lattice . .
complete meet
homomorphism . . .
composition . . . .
Condition I . . . .
. . . 75
pa
p;
ge
1
36
Condition II
67 converse . . . .
2 corresponding sections
6 deterministic
Travis matrix . .
6
diagonal . . . .
12
diagonal ...
12 restricted to .
4 disjoint sum ... .54,
6 distributive lattice
11 distributive triple
4 domain . . . .
7 dominate . . . .
7 dual . . . . .
epimorphism . . .
16
equivalence class
17 equivalence relation
5 FoulisHolland Theorem
greatest (element)
16
rsequence . . .
2
Fstate . . .
75
rstate space . .
page
horizontal sum .. .54,55
implements . . .. .24
infimum . . . . 4
irreducible . . .. .12
isomorphism . . .. 17
join . . . . 4
join dense . . . 7
join homomorphism .16
joinmeet homomorphism 16
largest (element) .. 4
lattice . . . . 5
least (element) . . 4
linearly ordered subset 4
lower bound . . . 4
meet . . . . 4
meet homomorphism . 16
modular lattice . 15
modular pair ... .15
monomorphism ... .17
one . . . . . 6
order filter ... .18
order homomorphism 16
order ideal . . .. .18
orthocomplementation 7
orthocomplemented lattice
orthocomplemented poset
orthogonal . . .
orthogonal family .
orthohomomorphism . .
orthomodular identity
orthomodular lattice
orthomodular poset .
partially ordered set .
partially orders . .
partial ordering . .
polynomial in M . . .
poset . . . . .
poset with zero and one
principal order filter
principal order ideal
principal section . .
proper order filter .
proper order ideal .
range . . . . .
reducible . . . .
reflexive . . . .
relation . . . .
representative . . .
page
9
7
9
9
16
9
9
9
3
3
3
13
3
6
18
18
19
18
18
1
12
2
1
3
section . . . .
simple (section) . .
smallest (element) .
subcomplete . ...
sublattice . . .
suborthomodular lattice
supremum . . . .
symmetric . . . .
page
19
56
4
5
5
9
4
2
transitive . . .
Travis matrix . .
trivial order filter
trivial order ideal
unit . . . .
upper bound .....
"weak" partial order
induced on P* by Z*
zero . . .
viii
page
2
83
18
18
6
4
83
6
CHAPTER 1
DEFINITIONS AND ELEMENTARY RESULTS
1. From Relations to Orthomodular Lattices
1.1.1 Notation. We assume that the reader has a
knowledge of the concepts of set theory (including the
various forms of the axiom of choice) and cardinal number
theory. The cardinal number of a set M is denoted by the
# M
symbol #M. The power set of a set M is denoted by 2 If
M = m, then 2 is sometimes written 2m.
The symbols E, c, and n are used in their usual
settheoretic sense. The complement of N in M is denoted
by MN. Finally, the symbol T.A.E. stands for "the
following statements are equivalent."
1.1.2 Definition. Let X be any set. We define XXX to
be the set [(x,y):x,yEX). By a relation on X we mean a
subset of XXX. If R is a relation then we call the
relation [(x,y):(y,x)ER) the converse of R and write it as
1
R1. {x:(x,y)ER) is called the domain of R and fy:(x,y)ER}
is called the range of R. Note that the domain of R equals
the range of R and the range of R equals the domain of
the range of R and the range of R equals the domain of
l
R1. Let M c X and xEX, then MR denotes
(y:(z,y)ER for some zEM) and xR denotes {x)R. (Here, as in
the sequel, relations operate on the right. In keeping
with this convention, most functions operate on the right;
however, in analytically flavored contexts functions may
operate on the left.) If R and S are two relations on X,
then we define the composition of R and S, denoted by RS,
to be the relation [(x,y):there exists zEX such that
(x,z)ER and (z,y)ES). By the diagonal of XXX, symbolized
by AX or simply A if there can be no confusion' as to what
set is under consideration, we mean ((x,x):xEX). If Y c X,
then the diagonal of XXX restricted to Y, written AIy, is
defined to be [(x,x):xEYI.
Let R be a relation on X. Then R is said to be
l
reflexive if A c R; R is said to be symmetric if R c R; R
is said to be antisymmetric if (x,y)ER and (y,x)ER imply
x = y; R is said to be transitive if RR c R. We sometimes
2 3
write RR as R (RR)R as R etc.
1.1.3 Remark. (1) Any relation of the form RUR is a
symmetric relation.
(2) A is a symmetric relation.
(3) The union of two symmetric relations is again a
symmetric relation.
(4) Let R a in some index set I, and T be relations
on X. Then ( U R )T = U (R T) and
aEI a aE a
T(U R ) = U (TR).
a6I aEI a
The proof follows immediately from the
definitions.
1.1.4 Definition. If R is a reflexive, symmetric, and
transitive relation on X, then R is called an equivalence
relation on X. If R is an equivalence relation, then a set
of the form xR is called an equivalence class, the set of
all such equivalence classes is denoted by X/R, and an
element of such an equivalence class is called a
representative of the equivalence class. The following
statement is an immediate consequence of the above
definition: If R is an equivalence relation, then x is a
representative of the equivalence class yR if and only if
xR = yR.
If R is reflexive, antisymmetric, and transitive
on X, then R is called a partial ordering on X; in this
case (X,R) is called a partially ordered set or simply a
poset.
If s is a relation on the set P such that (P,s)
is a poset, then we say that S partially orders P. If
x,yEP and x S y, then y is said to dominate x, x is said to
be dominated by y. If at least one of x 9 y, y s x holds,
then x and y are said to be comparable. If x and y are
comparable for all x,y in a subset M of P, then M is called
a linearly ordered subset of P or a chain in P. If x is
dominated by y and x X y, then we sometimes write x < y.
Let M c P; if uEP is such that xEM implies that x & u, then
u is called an upper bound for M; if VEP is such that xEM
implies that v & x, then v is called a lower bound for M.
The set of all upper bounds for M is denoted by
U(M), i.e.,
U(M) = (zEP:x s z for all xEMJ.
The set of all lower bounds for M is denoted by
T(M), i.e.,
T(M) = fzEP:z S x for all xEM].
An element z6P is called the greatest (largest)
element of a subset M of P if zEM and zEU(M). An element
zEP is called the least (smallest) element of a subset M of
P if zEM and zET(M). An immediate consequence of the fact
that s is antisymmetric is the fact that the greatest and
least elements of a subset M of P are unique (if they
exist). An element z of P is called the supremum (join) of
the subset M of P, written sup M, if z is an upper bound
for M, and, whenever u is an upper bound for M, z E u. An
element z of P is called the infimum (meet) of the subset M
of P, written inf M, if z is a lower bound for M, and,
whenever v is a lower bound for M, v S z.
A poset (P,t) is called a lattice in case, for
each nonempty finite subset M of P, there exist z,wEP such
that z = sup M and w = inf M. In case M = (x,y), then
sup M is sometimes written xvy and inf M is sometimes
written xAy. Also if M = IxI, ..., xn], then sup M and
inf M are sometimes written xl1...vxn and x A...AXno
respectively. It may be shown that sup and inf satisfy the
usual (generalized) associative and commutative laws.
The statement that the supremum (resp., infimum)
of a subset M of P exists is clearly equivalent to the
statement that U(M) has a least element (resp., T(M) has a
greatest element). To say that z = sup M (resp.,
w = inf M) is to say that U(M) = U((z}) (resp.,
T(M) = T([w))).
A poset (P,g) is called a complete lattice if,
for every M c P, there exist z,wEP such that z = sup M and
w = inf M; if such is the case we say simply that sup M and
inf M exist.
A subset M of a lattice (L,A) is called a
sublattice of L in case (M,IM) is a lattice. A sublattice
M of the lattice L is called subcomplete in case the
following obtains: if N c M and sup N exists as computed
in L, then sup N exists as computed in M, the two are
equal, and the common value is in M. A sublattice M of a
lattice L may be (1) complete but not subcomplete, (2)
subcomplete but not complete, (3) neither, or (4) both.
Let L be the lattice of all subspaces of an infinite dimen
sional Hilbert space H (with settheoretic inclusion as the
partial order), and let M be the sublattice of L consisting
of all closed subspaces of H. Then M is an example of (1).
Any lattice which is not complete, considered as a
sublattice of itself, is an example of (2). The lattice of
all finite subsets of an infinite set X, considered as a
sublattice of the power set of X, is an example of (3).
Any complete lattice, considered as a sublattice of itself,
is an example of (4).
An element 0 (resp., 1) of a poset P with the
property that U(0) = P (resp., T(1) = P) is called the zero
(resp., unit) of P. The uniqueness of the zero (resp.,
unit), if it exists, is an immediate consequence of the
fact that the partial ordering is antisymmetric. The unit
is sometimes called the one of the poset. If there exists
a zero and a one in P, then P is called a post with zero
and one. In a poset P with zero and one, P(0,x) denotes
(zEP:0 S z s xJ and P(x,l) denotes IzEP:x & z s 1).
If P contains a zero element O, then an element x
in P is said to be an atom of P in case 0 < x and if
0 < y < x for some yEP, then y = 0 or y = x. If P contains
a one element 1, then an element x in P is said to be a
coatom of P in case x < 1, and if x : y S 1 for some yEP,
then y = x or y = 1. P is said to be atomic in case every
nonzero element of P dominates an atom of P. If every
nonzero element of P is the join of the atoms it
dominates, then the atoms of P are said to be loin dense.
If x is an element of a poset P with zero and
one, then an element y of P is called a complement of x in
P in case xvy exists, xAy exists, xvy = 1, and xAy = 0. If
every element of a poset P has a complement in P, then P is
called a complemented poset. If P is a complemented poset
and if, moreover, there is a mapping ':P > P such that
(1) x 9 y implies y' s x',
(2) x'' = x (where, by definition, x'' = (x')'), and
(3) x' is a complement for x,
then P is called an orthocomplemented poset. ':P > P is
said to be an orthocomplementation on P.
1.1.5 Notation. We reserve the right to use any of the
symbols P, (P,t), or (P,Z,') to represent the (orthocom
plemented) poset P. By judicious utilization of this
standard abuse of terminology, we will single out the
salient feature of P under discussion, distinguish between
two such structures if necessary, and maintain a minimum of
notational overweight.
1.1.6 Definition. Because of the symmetry in the
definitions of supremum and infimum, zero and one, we have
the following principle of duality for posets with zero and
one:
If, in any valid statement which holds for all
posets with zero and one, we interchange the symbols for
supremum and infimum, interchange zero and one, and reverse
all inequalities, then we obtain another valid statement
(called the dual of the original statement) which holds for
all posets with zero and one.
Similar principles of duality hold for general
posets and for orthocomplemented posets.
1.1.7 Lemma. Let P be an orthocomplemented poset, then
P satisfies the Generalized DeMorgan Laws, i.e., for
x EP (aEI), (1) if sup (xa:a6I) or inf [x':alEI exists in
P, then they both exist and (sup {x :aEI])'= inf (x':aEI);
and (2) if inf (x :aEIj or sup (x:alEI} exists in P, then
they both exist and (inf (x :aEI})' = sup ({x:aEI].
Proof. If sup (x :aEI) exists and equals x, then
x x for all aEI, so that x' x' for all aEI. If y is
any other lower bound for x', then y' x for all aEI,
y' ; x, and finally y & x'. Hence x' = inf {x':aEI).
Similarly, if inf (x':aEI exists and equals x, then
x' = sup (x :aEI). Hence (1) is valid. (2) follows from
(1) by duality.
1.1.8 Definition. Two elements x,y of an ortho
complemented poset are said to be orthogonal, written
x _L y, in case x t y'. Note that if x __ y if and only if
y x. A family (x :aEA} of elements of P is said to be
an orthogonal family in case a 7 B implies that x xg.
An orthocomplemented poset P is called an orthomodular
post in case P satisfies the following two properties:
(1) if x,yEP are such that x y, then xvy exists
in P, and
(2) for x,yEP, x s y implies y = xv(y'vx)'.
The latter condition is called the orthomodular identity,
abbreviated OMI.
An orthomodular (resp., orthocomplemented) poset
which is a lattice is called an orthomodular (resp., ortho
complemented) lattice. A sublattice L1 of an orthomodular
lattice L is said to be a suborthomodular lattice of L,
written L1 < L, in case the restriction of the orthocom
plementation on L to L makes L1 an orthomodular lattice.
In case L1 < L and L1 L, we sometimes write L1 S L.
2. Standard Results in the Theory
of Orthomodular Lattices
1.2.1 Theorem. Let L be an orthocomplemented lattice.
Then T.A.E.
(1) L is an orthomodular lattice, i.e., L satisfies
the OMI,
e a f implies f = ev(fAe'),
for all e,fEL,
(2) L satisfies the dual orthomodular identity,
abbreviated DOMI,
e f implies e = fA(evf'),
for all e,fEL,
(3) *eCf implies fCe, for all e,frL,
(4) e < f and fAe' = 0 implies e = f,for all e,fEL,
(5) L does not contain a sublattice of the form
1
Figure 1
Proof. The equivalence of (1), (2), and (3) is
proved in [4, Theorem 1, p. 68]. The equivalence of (4)
and (5) is obvious. Moreover, it is clear that (1) implies
(4). Hence it suffices to prove that (5) implies (1).
Suppose that (5) is valid but that (1) is
invalid. Then there exist a,fEL such that a f and
f > av(fAa'). Let e = av(faa'). We claim that
{0,l,e,e',f,f'} is a sublattice of L of the type given in
Figure 1. For, f a e implies fAa' a eAa'. Moreover,
e = av(fAa') implies e 2 fAa'; hence eAa' a (fAa')Aa' = fAa'
and consequently fAa' = eAa'. Since
f've e ; eAa' = fAa' f fAe' = (f've)', it follows that
f've = 1, and hence fAe' = 0. Therefore fve' = 1 and
eAf' = 0. But fAf' = eAe' = 0 and fvf' = eve' = 1. Hence
(0,l,e,e',f,f'] is a sublattice of L of the type given in
Figure 1 which contradicts the assumption that (5) is
valid, i.e., L contains no such sublattice. Consequently
(5) implies (1).
1.2.2 Corollary. The atoms of an atomic orthomodular
lattice are join dense.
Proof. The result follows from an application of
part (4) of Theorem 1.2.1.
1.2.3 Definition. Let L be an orthomodular lattice.
Corresponding to every element fEL, we define a mapping
pf:L > L as follows:
For eEL, ecpf = (eVf')Af.
For e,fEL, we say that e commutes with f if and only if
ecf = eAf in which case we write eCf. Also, for any subset
M of L we define
C(M) = {eEL:eCf for all fEM).
By CC(M) we mean C(C(M)). The set C(L) is called the
center of L. Note that (0,1) c C(L) c L always holds. If
(0,1) = C(L), then L is said to be irreducible; if
c
(0,1) f C(L), then L is said to be reducible. If
C(L) = L, then L is said to be Boolean.
1.2.4 Theorem. Let L be an orthomodular lattice, and
let M,N c L.
(1) C(M) is a subcomplete suborthomodular lattice
of L.
(2) M c N implies C(N) c C(M).
(3) M c C(C(M)).
(4) C(M) = C(C(C(M))).
(5) M = C(C(M)) if and only if there exists N c L
such that M = C(N).
(6) Let M be a suborthomodular lattice of L.
Then T.A.E.
(a) M is a Boolean lattice,
(b) M c C(M),
(c) C(C(M)) C C(M).
(7) If M c C(M), then C(C(M)) is a subcomplete
Boolean suborthomodular lattice of L.
Proof. (1) is essentially proved in
[2, Lemma 3, p. 67). (2) and (3) follow immediately from
Definition 1.2.3; (4), (5), and (6) are immediate conse
quences of (2) and (3). (7) follows from (1), (4), and (6).
1.2.5 Definition. Let L be an orthomodular lattice,
and let M c L. By a polynomial in M, written Pol(M), is
meant any finite combination of supreme and infima of
elements of M or primes of such elements, or any finite
combination of supreme and infima of such combinations of
elements of M or primes of such combinations of elements of
M, etc.
It is to be emphasized that, although a poly
nomial in M may be represented in different ways as
combinations of elements of M or the primes of such
elements, in any given representation only a finite number
of elements of M appear and that the symbols v, A, and
appear only a finite number of times.
The following proposition contains the elementary
properties of polynomials which will be utilized in the
sequel.
1.2.6 Proposition. Let L be an orthomodular lattice,
and let M,N c L.
(1) If M c N, then Pol(M) c Pol(N).
(2) Pol(Pol(M)) = Pol(M).
(3) If M c Pol(N), then Pol(M) c Pol(N).
Proof. These results follow immediately from
Definition 1.2.5.
1.2.7 Proposition. Let L be an orthomodular lattice,
and let e,f,gEL.
(1) T.A.E.
(a) eCf,
(b) fCe,
(c) e'Cf.
(2) If e Cf for all aEI, then (sup (e :aEI})Cf.
(3) If e f, then eCf.
(4) eEC(L) if and only if e has exactly one
complement in L.
Proof. (1), (2), and (3) are proved in
[2, Lemmata 1, 2, and 3, p. 67]. (4) is proved in
[4, Corollary to Theorem 4, p. 71].
1.2.8 Theorem (FoulisHolland Theorem). Let L be an
orthomodular lattice, and let e,f,gEL. If any two of the
three relations eCf, eCg, fCg hold, then
(eVf)Ag = (eAg)v(fAg) and (eAf)vg = (evg)A(fvg).
Proof. The proof is found in [2, Theorem 5,
p. 68] and in [4, Theorem 3, p. 69].
An immediate consequence of Proposition 1.2.7 and
the FoulisHolland Theorem is the following
1.2.9 Corollary. Let L be an orthomodular lattice, let
M c L be any family of mutually commuting elements of L,
let eEL be such that eCf for all fEM, and let pEPol(M).
Then eCp.
Proof. poe = (pve')Ae = (pAe)v(e'Ae) = pAe.
1.2.10 Definition. Let L be a lattice, and let a,b,cEL.
Then we say that (a,b,c) is a distributive triple and we
write D(a,b,c) in case (avb)Ac = (aAc)V(bAc). We say that
(b,c) is a modular pair and we write M(b,c) in case
D(c,a,b) holds whenever c 9 b. L is said to be a
distributive (resp., modular) lattice in case D(a,b,c)
holds for all a,b,cEL (resp., M(b,c) holds for all b,cEL).
1.2.11 Remark. An orthomodular lattice is a Boolean
lattice if and only if it is a distributive lattice.
Proof. This is an immediate consequence of the
FoulisHolland Theorem.
1.2.12 Lemma. Let L be an orthomodular lattice, and let
M c L be any family of mutually commuting elements of L.
Then Pol(M) is a Boolean suborthomodular lattice of L.
Proof. Let p,q,rEPol(M). Apply Proposition 1.2.7
to the components of p, q, and r to obtain that p, q, and r
commute. Then apply the FoulisHolland Theorem to obtain
D(p,q,r).
1.2.13 Lemma. Let B be a Boolean lattice, let xEB, and
let a be an atom of B. Then either a s x or x s a'.
Proof. a = aAl = aA(xvx') = (aAx)v(aAx'). Since
a is an atom of B, 0 < aAx < a and 0 s aAx' s a. If both
aAx = 0 and aAx' = 0, then a = 0 which is a contradiction.
Hence aAx = a or aAx' = a. If both obtain then a S x and
a x', hence a = 0 which is a contradiction. Hence
exactly one of a x, x s a' obtains.
3. Morphisms, Ideals, Filters,and Sections
1.3.1 Definition. Let P and Q be posets. Let M c P
and N c Q. Then a mapping f from M into N is said to be
an order homomorphism in case x,yEM and x s y imply that
xf t yf. If f has the property that, for all x,yEM, xvy
exists in M implies xfvyf exists in N and (xvy)f = xfvyf,
then f is called a join homomorphism. A meet homomorphism
is defined dually. If f is both a join homomorphism and a
meet homomorphism, then f is said to be a joinmeet
homomorphism.
If M and N are suborthocomplemented posets of P
and Q, respectively, then a function f:M > N is said to
be an orthohomomorphism in case x'f = (xf)'. If f has the
property that, for all K c M such that sup K exists and is
in M, sup (Kf) exists and (sup K)f = sup (Kf), then f is
said to be a complete join homomorphism. A complete meet
homomorphism is defined dually. A homomorphism which is
both a complete join homomorphism and a complete meet
homomorphism is said to be a complete joinmeet
homomorphism.
If f is a homomorphism of any type, then f is
1
said to be a monomorphism from M to N in case f :Mf > M
is a homomorphism of the same type as f, f is said to be an
epimorphism from M to N in case Mf = N, and f is said to be
an isomorphism from M to N in case f is both a monomorphism
and an epimorphism.
1.3.2 Lemma. Let (LLl1) and (L2, 2) be orthomodular
lattices, let S, and S2 be suborthomodular lattices of L
and L2, respectively, and let 9:S1 S2 be an order ortho
isomorphism. Then 9 is a joinmeet orthoisomorphism. If,
in addition to the above hypotheses, L1 and L2 are complete
lattices and S1 and S2 are subcomplete sublattices, then 9
is a complete joinmeet orthoisomorphism.
Proof. We assume that Sl and S2 are subcomplete
and prove that 8 is a complete joinmeet orthoisomorphism.
The proof of the other statement is merely a simplification
of the proof given.
Let M c Sl, m = sup1 M, and n = sup2 (MG) (where
supi indicates the supremum in Li). Since S1 and S2 are
subcomplete, mESL and nES2; hence there exists wES1 such
that w8 = n. We must show that w = m. From x 5l m for all
xfM, it follows that x8 <2 m6 for all x8EM9; hence
suP2 (MS) = n = we <2 me, and therefore w 9l m. But from
x8 S n = we for all xARM9, it follows that x sl w for all
xEM; hence sup1 M = m i1 w. Therefore w = m and conse
quently 8 is a complete join homomorphism. By a dual
argument, 9 is a complete meet homomorphism. A similar
argument proves that 1 has these properties; the result
follows.
1.3.3 Definition. A subset I of an orthocomplemented
poset P is said to be an order ideal of P if and only if
mEI and n < m imply nEI. A subset F of P is said to be an
order filter of P if and only if mEF and m S n imply nEF.
If I (resp., F) is an order ideal (resp., order
filter) on P with I # P (resp., F X L) then I (resp., F) is
said to be a proper order ideal (resp., proper order
filter) on P. If I = [0,1) or I = P (resp., F = {0,1) or
F = P), then I (resp., F) is said to be a trivial order
ideal (resp., trivial order filter). An order ideal I
(resp., order filter F) is said to be a principal order
ideal (resp., principal order filter) in case there exists
xEP such.that I = P(0,x) (resp., F = P(x,l)).
Note. Let I & L. I is an order ideal of P if
and only if the set F = (xEL:x'EI) is an order filter of P.
The following terminology is not standard.
1.3.4 Definition. A subset S of the orthocomplemented
poset P is said to be a section of P in case S = IUF where
I is an order ideal of L, F is an order filter of L, and
xEI if and only if x'EF. If S = L(x,l)UL(0,x'), then S is
called a principal section of P and S is denoted by S x
1.3.5 Remark. Let P be an orthocomplemented poset.
Then every section of P is the union of principal sections
of P.
Proof. Let S be a section of P, then S = IUF for
some order ideal I and some order filter F. Then
S = U[Sx:xEI).
1.3.6 Lemma. Let P be an orthocomplemented poset, and
let S be a section of P. Then S is closed under the ortho
complementation of P.
Proof. By Remark 1.3.5 we need only show that
any principal section S is closed under the orthocomple
mentation of P. We may assume x $ 0. If yESx, then either
x s y or y x'. If x y, then y' : x' so that y'ES If
y s x', then x & y' so that y'ES Hence S is closed
under the orthocomplementation of P.
1.3.7 Remark. If Sx is a principal section of an
orthomodular lattice L, then S is a suborthomodular
lattice of L.
20
Proof. By Lemma 1.1.7 we need only prove that
the join (in L) of any two elements in Sx is again in Sx.
But this follows immediately from the definition of S
x
CHAPTER 2
THE PASTE JOB
1. New Lattices from Old
2.1.1 Convention. Throughout this chapter we assume
4t +
that (Ll,:, #) and (L2,2, ) are two disjoint orthomodular
lattices, and that S. c L. (i = 1, 2) are such that:
(1) [0,1) c Si 4 Li (i = 1, 2);
(2) S. is closed under the orthocomplementation of
Li (i = 1, 2);
(3) there exists e:S1 > S2 such that 8 is an order J
orthoisomorphism.
The ambiguity of notation generated by using the
symbol S1 for the principal section and for the subset of
L1 will be resolved by the context in which the symbol
appears. Unless otherwise stated S1 will denote the subset
of L1 given in the above convention.
2.1.2 Definition. (1) Let L0 = L UL2.
(2) Let P1 = ((x,y)EL0XLO:y = x)].
(3) Let A = {(x,x):xEL)}.
(4) Let P = APUP1
(4) Let P = AUPlUPI
1 1*
2.1.3 Proposition. P is an equivalence relation on L0.
Proof. (i) P is reflexive since A c P.
(ii) P is symmetric by Remark 1.1.3.
(iii) To show that P is transitive, we must prove
that PP c P. We first note that PIP1 = since the range
of P1 is disjoint from the domain of Pl. For a similar
reason P 1 = (. We also note that PP11 = AIS and that
PI Pl = A S ; both identities are immediate from the
2
definition of PV. We now compute:
1 1
PP = (AUPLUP )(AUP1UP1
= A(AUUPP )UPP(AUPlUPI1 )u1 (PAUPlI1
1 1 1 1 1
= AAUAP U(API )UP1AUPP 1UP1P1 UP1 AUP 1PUPP 1
1 1 1 1 1 1
1 1 1 u1 1p1 UP1 1 1uplp1
= AUP UP1UPUP P UPP 1 UPI UP 1pUPP1
1 1 1 1 1
= AUP1UP1 UPIPUPIP1 UP1I P1UPp1 p1
= (AUPUP l)UVUAlS UAIS Ub
1 2
1
= AUPIUP1 P.
Hence P is transitive.
2.1.4 Definition. (1) Let L = L0/P.
(2) Let R1 = (([x],[y])ELXL:there exist x1E[x] anc
ylE[y] such that xl 1l Y13}
(3) Let R2 = (([x],[y])ELXL:there exist x2E[x] anc
y2E[y] such that x2 2 Y2).
(4) Let R = {([x],[y])ELXL:there exists [z]EL such
that ([x],[z])ER1UR2 and ([z],[y])ER1UR21.
d
d
2.1.5 Remark. R = (R UR 2)
Proof. This is an immediate consequence of
Definition 2.1.4 part (4) and the definition of (RIUR2)
2.1.6 Lemma. (i) If y = x9, then xeS1 c L1 and
yES2 c L2.
(ii) If y = xG and z = x6, then y = z.
(iii) If y = x9 and y = z8, then x = z.
(iv) If [x]EL, then either [x] is a singleton set
whose only element is in L1S1 or in L2S2, or a
doubleton set consisting of an element xlES1 and
an element x2ES2 such that xl = x2.
(v) If xl,x2E[x], yly2E[y], x19 = x2, and yl1 = y2'
then the following are equivalent:
(a) xl 1 Y'
(b) x2 &2 Y2'
(c) y 1 x ,
+ +
(d) y2 2 x2'
Proof. (i) is clear because the domain of e is
contained in S1 which in turn is a subset of L1 and the
range of 0 is contained in S2 which in turn is a subset of
L2'
(ii) is clear because 0 is, among other things,
a function.
(iii) is clear because 0 is a monomorphism.
(iv) follows immediately from the fact that
1
P = AUPUP1
In (v), (a) is equivalent to (c) since is an
orthocomplementation for LI, and (b) is equivalent to (d)
+ .
since is an orthocomplementation for L2. Moreover (a) is
equivalent to (b) since 0 is an order isomorphism. Hence
all four statements are equivalent.
2.1.7 Notation. If [z]EL is such that ([x],[z])ER1UR2
and ([z],[y])ERlUR2, then we say that [z] implements
([x],[y])ER and we write [z]:[x]R[y].
For i,jEl1,2), if [z]EL is such that ([x],[z])ERi
and ([z],[y])ERj, then we write [z]:[x]R.iR[y] and we say
that [z] implements ([x],[y])ER.R..
In what follows [x] = {xl), [x] = (x2], and
[x] = lxl,x2) mean, respectively, that [x] = {xl c L Sa ,
[x] = (x2] c L2S2, and [x] = [(x,x2} where x1ES1, x2ES2,
and xl1 = x2. We will freely write [x1] for [x] whenever
xlE[x], and [x2] for [x] whenever x2E[x]. A subscript 1 or
2 will always denote the fact that the element in question
is in L1 or in L2, respectively. The home of unsubscripted
elements either will be made explicit or will be left
conveniently ambiguous if no confusion is generated by the
ambiguity.
2.1.8 Lemma. (1) If [z]:[x]R[y], [x] = (x1], and
[y] = [y2], then [z] = fzl,z2) and x 1 I zl'
z2 ;2 Y2'
(2) If [x]R[y], [x] = fxl,x2), and [y] = [ylY2],
then [x]:[x]R[y] and [y]:[x]R[y].
Proof. Ad (1). One of [z] = (zl], [z] = {z2'}
or [z] = (z1,z2] must obtain. If [z] = (zl), then
([z],[y])ER1UR2 which is impossible. If [z] = (z2], then
([x],[z])ERIUR2 which is impossible. Hence [z] = (zl,z2),
xl S1 Z1, and z2 2 Y2'
Ad (2). By the use of the order homomorphism 8,
it is easily shown that xl 1 y1 and x2 '2 Y2. The result
follows.
2.1.9 Proposition. R is a reflexive, antisymmetric,
and transitive relation on L.
Proof. (1) R is reflexive since [x]:[x]R[x].
(2) To prove that R is antisymmetric, assume
that [z]:[x]R[y] and [w]:[y]R[x]. We must show that
[x] = [y]. Note that if [x], [y], [z], and [w] all have
representations in L1, then the following obtains:
(I) 1 Z1 Z 1 y1 l w1 1 l
Hence x, = yl and [x] = [y]. Note also that if [x], [y],
[z], and [w] all have representations in L2, then the
following obtains:
(II) x2 s2 z 2 2 2 Y22 w22 x2
Hence x2 = y2 and [x] = [y].
One and only one of the following cases obtains:
Case (1) [x]
Case (2) [x]
Case (3) [x]
Case (4) [x]
Case (5) [x]
Case (6) [x]
Case (7) [x]
Case (8) [x]
Case (9) [x]
Cases (1) z
are Cases (3), (6),
= (x)l
= (xl)
= (xl)
= (x2)
= (x2)
= (x2)
= (xl,x23
= 1xl,x2)
= (Xl,x2]
nd (5), (2)
(7), and (8)
and
and
and
and
and
and
and
and
and
and (4)
[y] = IYl),
[y] = (y2),
[y] = (ylY2),
[y] = (y],
[y] = (y2),
[y] = (ylY21'
[y] = C(y)'
[y] = (y2),
[y] = [ylY2.
are symmetric, as
.Accordingly we only
consider Cases (1), (2), (3), and (9).
In Case (1), (I) obtains; hence [x] = [y].
In Case (2), by Lemma 2.1.8 part (1), we have
[z] = z11,22), [w] = [lW2)}, and x1 S1 Zl z2 2 2 Y2'
Y2 :2 w2' W 1 xl Hence wl 1 zl and z2 a2 w2. But,
since 0 is an order homomorphism, z2 S2 w2 = w 1 <2 z18 =22'
and so z2 = w2. Moreover, since 0 is a monomorphism,
z1 = w1. Hence xl 1 Zl = wl L xl and y2 92 w2 = z2 92 Y2
imply that xl = z and y2 = z2. Consequently
[x] = [xl] = [z1] = [z2] = [y2] = [y] which is the desired
result.
In Case (3), (I) also obtains which yields a con
tradiction since no singleton equivalence class can equal a
doubleton equivalence class. Hence Case (3) cannot occur.
In Case (9) we may assume that [z] = [w] = [x]
(cf. Lemma 2.1.8 part (2)). Hence (I) occurs and [x]= [y].
Consequently R is antisymmetric.
(3) To prove that R is transitive, we must show
that
(III) RR c R.
2
But R = (R UR22) by Remark 2.1.5, and hence we may compute
RR as follows:
RR = R(R UR2)2
2 2
= R(R2UR R2UR2R UR2)
2 2
= RR2URR1R2URR2R1URR2
21 2 2 2
= (RIUR2) 2R1U(R UR2) 2R1R2U( 2 RRRUR2 2R (R1UR2) R2
S2 2 2 2 2
= (R1UR1R2UR2RUR2) RU (R1UR1R2UR2R UR2)R1R2
112 21 21 11 2 1 22 2
U (R2UR1R2UR2RIUR2) R2RU (R2UR R2UR2RUR2)R2
4 2 2 22 3 2 2 2
= (RIUR R2R1UR2R UR2R1)U(R R2U(R R2) UR2RIR2UR2R1R2)
2 2 23 22 3 24
U (RR2RIUR1R2RIU (R2R) 2UR2R1)U(R1R2UR1R2UR2R R2UR) .
Hence (III) is equivalent to
4 2 3 2 2 3 2 2 2
(IV) R1UR R2R1UR2R1UR2R2UR1R2U (R R2) UR2R1R2UR2R1R2
2 2 2 3 2 2 3 2 4
UR1R2R1UR R2RU (R2R1) UR2R1UR2R2UR1R3UR2RR2U2R2
2 2
c R2UR1R2UR2R UR2.
To prove (IV) we consider 16 cases corresponding
to the 16 sets exhibited on the left hand side of (IV). In
each case we will show that the set in question is
contained in one of the sets on the right hand side of (IV).
Case (1). If ([x],[y]))ER, then there exist
[p],[q],[r]EL such that
([x],[p]),([p],[q]),([q],[r]),([r],[y])ER1. Hence there
exist xlE[x], plE[p], qlE[q], rlE[r], and ylE[y] such that
xi 1l pi 1 q1 1 r 1 Y which implies x 1 1q, Y1
Therefore ([x],[q]),([q],[y])ER1, [q]:[x]R2[y], and
([x],[y])ER2.
Case (2). If ([x],[y])ER1R2R, then there exist
[p],[q],[r]EL such that ([x],[p])(R1, ([p],[q])ER2, and
([q],[r]),([r],[y])ER.. By definition of R1 and R2,
[P] = (p',p2}' [q] ={qlq2], x1 '1 p1' P2 P 2 q2, and
q91 1 rl 1 YI Hence (via 0) p, 1 ,1 and therefore
xl1 Sp 1 1 1 q~i 1 rl1 ~ Yl which implies that [p]:[x]R2[y],
i.e., ([x],[y])R21.
Case (3). If ([x],[y])ER2R3, then there exist
[p],[q],[r]EL such that ([x],[p])ER2 and
([p1,[q]),([q],[r]),([r],[y])ER1. Hence [p] = fplP2]'
x2 &2 p2, and p, &l q1 11 r1 11 y1. Therefore
[p]:[x]R2Rl[y], i.e., ([x],[y])ER2R1.
22
Case (4). If ([x],[y])ER2R1, then there exist
[p],[q],[r]EL such that ([x],[p]),([p],[q])ER2 and
([q],[r]),([r],[y])ER Hence [q] = (ql,q2,
x2 2 P2 '2 q2, and q1 !1 rl 1 y which implies that
[ql:[x]R2R[y].
By arguments similar to those used in Cases (1) 
(4) we may conclude the following:
Case (5). If ([x],[y])ER R2, then there exists
[r]EL such that [r]:[x]R1R2[y].
Case (6). If ([x],[y])E(R1R2)2, then there
exists [r]EL such that [r]:[x]R1R2[y].
Case (7). If ([x],[y])ER2R2R2, then there exis
[r]EL such that [r]:[x]RRy].
Case (8). If ([x],[y])ER2R1R2, then there exis
[r]EL such that [r]:[x]R[y].
For n = 1, 2, .... 8. Case 16(nl) is obtained
from Case n by interchanging the subscripts 1 and 2.
Consequently R is transitive.
ts
ts
2.1.10 Notation. Since, by the preceding lemma, R is a
partial ordering on L, we write [x] A [y] whenever
([x],[y])ER. In particular, we will henceforth write
[z]:[x] S [y] for [z]:[x]R[y].
Moreover, in what follows, when considering a
relation such as [z]:[x] s [y], the statement "we may
assume that there exist xl6[x], zlE[z] such that xI sL1 Z"
will sometimes be abbreviated to the statement "we may
assume x 1 1 1 z"; also in case analyses the abbreviated
form "xI !1 z1" means "there exist x E[x], z E[z] such that
x1 1 zl*
2.1.11 Definition. Let 0. (resp., 1i) be the zero (resp.,
unit) element of Li (i = 1. 2). Define [0] to be [01]
and [1] to be [11]. Define ':L > L by the following:
l[x= I[x4] if there exists x1EL1 such that x1E[x].
[x]=
[x2] if there exists x2EL2 such that x2E[x].
2.1.12 Notation. We henceforth write 0. and 1.
1 1
(i = 1, 2) as 0 and 1, respectively. When euphony demands,
the notation {0,1)i will be utilized to denote {Oi,li]. In
other occurrences, if a distinction is necessary it will be
clear from the context.
2.1.13 Lemma. [0] and [1] are the zero and unit
elements, respectively, of L.
Proof. If [x]EL, then [x] = [xl] or [x] = [x2].
If [x] = [xl], then [0] a [x] since 0 '1 xl; if [x] = [x2].
we may similarly conclude that [0] < [x]. Hence [0] is the
zero element of L. Dually, [1] is the unit element of L.
2.1.14 Proposition. ':L > L is a welldefined ortho
complementation for the poset L.
Proof. ':L > L is welldefined. For, if
[x] = (x1}, or if [x] = (x2], then there is no ambiguity in
the definition of [x]' which is (x#] or fx +, respectively.
If [x] = {xlx2}, then, since x 9 = x, [x]' is unambiguous.
Now suppose that [x] S [a] and [x]' s [a].
Assume [z]:[x] [a] and [w]:[x]' a [a]. we may assume
that xl 1 zl and consider only the following eight
possibilities:
Case
Case
Case
.Case
Case
Case
Case
Case
x 1 r
[a] =
Xl 1
[a] =
xl 1l
hence
Xl 1 z1
[a] = [1]
(1) zl 1
(2) zl C
(3) z1 1~
(4) z1 ~i
(5) z2 i2
(6) z2 !2
(7) z2 '2
(8) z2 62
Ad (1).
Ad (2).
Z I al and
[1].
Ad (3).
z1 1 al and
[1].
al,
al,
al,
al,
a2,
a2,
a2,
a2,
Xl 1
[w] =
1
[x] =
x 4
1'1
x#1 ~1 Wl wl ~ al'
x I1 wl' w2 '2 a2'
x2 :2 W2' w1 ~l al'
x2 i2 2' w2 22 a2'
x S1 W1' W1 1 al'
X1 15 Wl w2 2 a2'
x2 22 w2 Wl S1 al'
x2 r2 w2' w2 2 a2'
al and x S1 al imply [a] =
(wl,w2] and [a] = (al,a2).
W1 1 al. Therefore al = 1
[x1,x2] and [w] = {w1,w2).
w1 1 al' Therefore al = 1
Ad (4). [x] = [xl,x2) and [a] = {al,a2).
[1].
Hence
and
Hence
and
Hence
1 al which implies x2 2 a2. But x2 2 a2 and
= 1. Consequently [a] = [1].
Ad (5). [a] a= (a,a2) and [z] = (z1,z2). Hen<
1 al and x#1 1 w al. Therefore al = 1 and
S.
Ad (6). [z] = [zl,z23 and [w] = {w1,w2}. Hence
xl i z1 which implies z1 ~1 x1 1 wl and therefore
ce
+
Z2 !2 w2 2 a2. But z2 2 a2 and hence a2 = i.
Consequently [a] = [1].
Ad (7). [x] = [(xX21}, [w] = (w1,w2},
[z] = [21,22], and [a] = fal,a2). Hence xL l Z1 '1 al and
x 1 W al. Therefore al = 1 and [a] = [1].
Ad (8). [x] = [Xl,x2) and [z] = (21,22]. Hence
x2 2 z2 a2, x2 2 w2 w 2 a2. Therefore a2 = 1 and
[a] = [1].
Hence [x]v[x]' exists and equals [1]. Dually,
[x]A[x]' exists and equals [0].
Clearly ':L > L is an involution since
#:L> L and +:L2> L2 are involutions. The verifi
cation that ':L > L is antiautomorphic is immediate.
For if [z]:[x] s [y], then we may assume xl 1 zl and we
need only check the cases z2 1 y1 and z22 2 y2. In the
former case xl1 S y implies y# al xl; hence
[y]' s [x]'. In the latter case y 2 4z2 and z# X
imply that [y]' [x]'. Hence ':L > L is an ortho
complementation for the poset L.
We now assume that the S. are proper sub
1
orthomodular lattices of L.. Then, by Lemma 1.3.2. the
order orthoisomorphism 9 becomes a joinmeet ortho
isomorphism. This allows us to make statements such as
the following: if [x] = [xl,x2} and [y] = (y1.Y2}, then
(x1 V1 Yl)8 = x2 v2 Y2 and (x1 A1 Y1) = x2 A2 Y2' i.e.,
[x V1 y1] = [x2 V2 y2], and [xl A1 y1] = [x A2 Y2 ]
2.1.15 Lemma. Let 0O,1]i Si < Li (i = 1, 2). Let
[x]ES and [y]ES so that [x] = (x1.x2] and [y] = (ylY21}.
Then the following obtains:
(1) [x]v[y] exists in L. and
[x]v[y] = [xl V1 y1] = [x2 V2 y2];
(2) [x]A[y] exists in L. and
[x]A[y] = [x1 1 yl] = [x A2 Y2].
Proof. Let [x] = [x1',x2 and [y] = {y1,Y2}.
Then, since Si is a sublattice of Li (i = 1, 2), x ,ylES1
implies xl v1 YlES1; x2'Y2ES2 implies x2 v2 Y2ES2.
Moreover (xl V1 y1)8 = x1a v2 y18 = x2 v2 Y2' hence
x1,'Y ~1 x1 V1 Yl implies [x],[y] ( [xl V1 y1] = [x2 V2 y2].
Now suppose that [x],[y] < [z]. The following
statements are valid regardless of the composition of the
elements of L which implement [x] : [z] and [y] [z].
If [z] = (zl] or if [z] = [(1,22], then xl A1 21
and y1 '1 z1. In this case it follows that x, v1 y1 l zl'
and hence [xl v1 yl] [z].
If [z] = (z2], then x2 2 z2 and Y2 2 z2. In
this case it follows that x2 v2 Y2 2 z2' and hence
[x2 V2 y2] < [z].
We have shown that, for [x],[y]ES,
[x],[y] C [Ix V1 y1] = [x2 V2 y2], and if [x],[y] i [z]
then [xl VI y1] S [z]. Hence we conclude that [x]v[y]
exists and equals [xl v1 y1].
2.1.16 Lemma. Let [0,1]i S Si < Li (i = 1, 2). Let
[x]ES and [y]hLS, then [x] = fx1,x2] and either [y] = [y1]
or [y] = (y2].
(1) If [y] = [yl), then [x]v[y]
[xv[y'] = [xI v1 y1];
(2) if [y] = [y21, then [x]v[y]
[x]v[y] = [x2 v2 Y2];
(3) if [y] = [y)1, then [x]A[y]
[x]A[y] = [x 1 A1 Y];
(4) if [y] = [y23, then [x]A[y]
[x]A[y] = [x2 A2 Y2].
exists in L and
exists in L and
exists in L and
exists in L and
Proof. Ad (1). [x] = [x,,x2) and
x1Y1 1 xl 1 V1 Y1 imply that [x],[y] C [xI v1 y1]. Now
assume [x],[y] a [z]. Let these inequalities be imple
mented by [u] and [v], respectively. If [z] = [zl) or if
[z] = (zl,z2), then xl ~1 z1 and yl '1 z1; hence
xl V1 Y ~1 Z1 which implies that [xl v, yl] s [z]. If
[z] = (z2}, then [v] = (v1,V2], y1 Y Vl' v2 2 z2, and
x2 <2 z2; xl v11 1 x1 V1 V1 and
(x1 V1 V1)8 = x2 v2 v2 '2 z2; hence [xl V1 yl] [z].
Hence [x]v[y] exists and equals [xl vI y1].
A similar argument proves (2). (3) and (4)
are dual.
2.1.17 Lemma. Let [0,1)i Si < Li (i = 1, 2).
(1) If [x] = (xl] and [y] = (yl), then [x]v[y] exists
in L and [x]v[y] = [xl VI y1].
(2) If [x] = (x2) and [y] = (y2), then [x]v[y] exists
in L and [x]v[y] = [x2 v2 y2].
(3) If [x] = (x1) and [y] = (y1), then [x]A[y] exists
in L and [x]A[y] = [x1 A1 y1].
(4) If [x] = (x2] and [y] = (y2), then [x]A[y] exists
in L and [x]A[y] = [x2 A2 y2].
Proof. We need only prove (1) since (2) follows
by symmetry; (3) and (4) are dual.
Clearly [x],[y] s [xl vI y1]. Now assume that
[x],[y] a [z] where [u] and [v] are the elements of L which,
respectively, implement the inequalities. Then both [u]
and [v] have representatives in uI and vl, respectively, in
L1. x1 I1 u1 and Y1 1 v1 imply xl VI yl al ul Vl vl. If
[z] = (zl) or if [z] = {zl,z2), then clearly
[u1 V1 V1]:[x1 V1 y1] < [z]. If [z] = [z22, then
[u] = [ulU2], [v] = v1,v2)], and u2 V2 v2 S2 z2; hence
[u1 v1 V1]:[x1 vI y1] [z]. Consequently, in this case,
[x]vly] exists and equals [xl v1 y1].
2.1.18 Definition. Let S1 < L1, and S2 < L2. Then we
say that S1 and S2 are corresponding sections of L1 and L2
if and only if there exist Mi c Si (i = 1, 2) such that
(1) M1B = M2'
(2) S1 = S # and S2 = S and
mEM1 mEM2 m
(3) 91Sm :Sm#aS(m )+ is an orthoisomorphism for each
mEM1.
Warning. Although a section need not, in general,
be a sublattice, in order for S1 and S2 to be corresponding
sections of L1 and L2, S1 and S2 must be suborthomodular
lattices of L1 and L2, respectively.
2.1.19 Lemma. Assume that S1 and S2 are corresponding
sections of L1 and L2. Under this assumption if [x],[y]EL
are such that [x] = (xl] and [y] = (y2), then [x] % [y].
Proof. Suppose the statement is false. Then
there exists [z] such that [z]:[x] < [y], [x] = {x)},
and [y] = (y23. It follows that [z] = [zl,z2), x1 L1 z'
z2 2 Y2, zlES1, and z2ES2. Hence z ESm# and z2ES(e)+
for some mEM1, so that either m# 5 zl or z1 !5 m. If
m# :1 zIl then (m) + 2 z 2 Y2. Therefore Y2ES(Me)+ S2
and hence [y]ES, which is a contradiction. If z1 ~1 m,
then xl 1 21 Z 1 m; and consequently x6ESm# C S i.e.,
[x]ES. This is a contradiction. Therefore no such [z]
exists and [x] [y].
2.1.20 Corollary. Assume that S1 and S2 are
corresponding sections of L1 and L2. Under this assumption
if [x],[y]EL are such that [x] = (xl1 and [y] = {y21, then
U([x],[y]) c S.
Proof. Suppose there exists [u]EU([x],[y]) such
that [u]OS. Then either [u] = {u1} or [u] = (u21. If
[u] = (ul), then [y] [u], contradicting Lemma 2.1.19. If
[u] = (u2), then [x] [u], again contradicting
Lemma 2.1.19. Hence no such [u] exists and U([x],[y]) c S.
2.1.21 Proposition. Let S1 and S2 be corresponding
sections of L1 and L2. Then L is an orthomodular poset.
Proof. By Proposition 2.1.14 L is an ortho
complemented poset. Hence we need only show that
(1) if [e],[f]EL and [e] __ [f], then [e]v[f] exists,
and
(2) if [e],[f]EL and [e] s [f], then
[f] = [e]v([f]'v[e])'.
Ad (1). If [e] _L [f], then [e] < [f]'. If
there exists elE[e], then by Lemma 2.1.19 there exists
flE[f], and by one of Lemmata 2.1.15, 2.1.16, or 2.1.17,
[e]v[f] exists (and equals ([e vl f1]). If [e] = (e2'}
then by Lemma 2.1.19, there exists f2E[f] and hence by one
of Lemmata 2.1.15, 2.1.16, or 2.1.17, [e]v[f] exists (and
equals [e2 V2 f2]).
Ad (2). Assume that [x] s [y]. We will prove
that [y] = [x]v([x]v[y]')'. We need only consider two
cases:
Case (i) [x] and [y] both have a representative in L1.
Case (ii) [x] = (xl) and [y] = (y2).
Ad (i). xl 1 yl and hence, by Lemmata 2.1.15,
2.1.16, and 2.1.17, since L1 satisfies the OMI,
[x]v([x]v[y]')' = [xl]v([xl]v[y ])' = [xl]V[x1 V1 y ]
= [X ]V[x# A1 y1] = [y1] = [y].
Ad (ii). By Lemma 2.1.19 this case cannot obtain.
Since the OMI is satisfied in all consistent
cases, L is an orthomodular poset.
2.1.22 Theorem. Assume that S1 and S2 are corresponding
sections of L1 and L2. Moreover, assume Li is complete and
that Si is a subcomplete suborthomodular lattice of Li
(i = 1, 2). Then L is an orthomodular lattice.
Proof. Since L is an orthomodular poset, to show
that L is an orthomodular lattice we need only show that
[x]v[y] exists for all [x],[y]EL. Let [x],[y]EL. Exactly
one of the following must occur:
(1) [x],[y]ES,
(2) [x]ES, [y]ELS (relabel [x] and [y]
if necessary),
(3) [x],[y]ELS.
Ad (1). By Lemma 2.1.15 [x]v[y] exists.
Ad (2). By Lemma 2.1.16 [x]v[y] exists.
Ad (3). There are four cases:
Case (i) [x] = {xL) and [y] = (yl},
Case (ii) [x] = (x1) and [y] = [y21,
Case (iii) [x] = [x2} and [y] = {yl,'
Case (iv) [x] = {x2) and [y] = (y2].
By symmetry we need only consider Cases (i) and
(ii).
Ad (i). By Lemma 2.1.17 [x]v[y] exists.
Ad (ii). First note that U([x],[y]) is nonempty
since it contains [1], and that, by Corollary 2.1.20,
U([x],[y]) c S. Let M1 = {zl:there exists [z]EU([x],[y])
such that [z] = (zl,z2]}. Now inf, M1 (as computed in LI)
exists since L is complete. But since S is subcomplete,
inf M,, as computed in S1, exists and equals infl Mi. Let
z() = infI M1. Let z(2) = z(1), let [z]0 = [z(1)
= [z(2)], and let M2 = fz2:there exists [z]EU([x],[y]) such
that [z] = {z1,z2]}. Then by Lemma 1.3.2
z(2) = inf2 (zle:zlEM1] = inf2 M2 = inf M2, as computed in
S2 since S2 is subcomplete.
We claim that U([z]0) = U([x],[y]).
Xl 1 infl ML = z(1) and y2 r2 inf2 M2 = z(2)
[x],[y] [)]0. Hence U([z]0) c U([x],[y]).
[z]EU([x],[y]), then by Corollary 2.1.20 [z]
z(1) 1 z1, and z(2) 2 z2; hence [z]0 < [z],
[z]EU([z]0). Therefore U([z]0) = U([x],[y]).
L is an orthomodular lattice.
2.1.23 Example. If
need not be a lattice.
be the Boolean lattice
For,
imply that
Moreover, if
= {z1,z2],
i.e.,
Consequently
Si is not a section of Li, then L
For example, for i = 1, 2, let Li
given in Figure 2.
Li = Xi
Figure 2
Let Si = [O,l,xi,x). Then Si is not a section of Li.
Form L as in Definition 2.1.4. (The quotes about ai, bi,
mi, and ni indicate that ale 4 a2, bl0 / b2, m10 Y m2, and
n18 0 n2 in violation of Notation 2.1.7.) Then
["al"]v["b2"] does not exist since
(["m"],["n2"],[l] ) = U(["al" ["b2"]) has no smallest element.
2.1.24 Example. The following example illustrates the
fact that if the suborthomodular lattice Si is not a sub
complete suborthomodular lattice of the complete ortho
modular lattice L. (i= 1, 2), then L need not be a lattice.
Let L be the power set of an infinite set M, and
let Sl be the suborthomodular lattice of L1 consisting of
all finite or cofinite subsets of L1. (Recall that a
cofinite subset of M is a subset of M whose complement in
M is finite.) Let L2 be a disjoint "copy" of L1. Then
there exists a natural orthoisomorphism c:L1"L2. Let
9 = cpS and let S2 = S A. Then e:SI S2 is an ortho
isomorphism mapping Sl onto S2. Clearly S1 and S2 are
corresponding sections of L1 and L2. Moreover, Li is
complete, but S. is not a subcomplete suborthomodular
lattice of L. (i = 1, 2). Form L as ir Definition 2.1.4.
To show that L is not a lattice, consider two
subsets X and Y of M which are neither finite nor cofinite
and such that XUY is not cofinite. Let x be the element
of Ll which corresponds to X and let y be the element of L2
which corresponds to Y. Then xCL1S1 and yEL2S2. Suppose
there is a [z]eL such that [z] = [x]v[y]. Then by
Corollary 2.1.20, [z]cs, and hence [z] = (zl,z2}. Let Z be
the subset of M corresponding to zI (and z2). Since
XUY c Z, Z is a cofinite subset of M. But, since XUY is
not cofinite, there exists wEZ(XUY) such that Z(w} = XUY.
Let V = Z([w and let v be the element of L1 which
corresponds to V. Then [v]EU([x],[y]) but [z] ; [v]. This
contradicts the fact that [z] = [x]v[y]. Hence [x]v[y]
does not exist and L is not a lattice.
The poset L given in the above example is of
interest for another reason. By Proposition 2.1.21 L is an
orthomodular poset. Hence L is an example of an ortho
modular poset which is not a lattice. The only other known
example of such a poset is a finite poset given by
M. F. Janowitz in [5].
2.1.25 Example.
Hasse diagrams:
S
L
a b
0
Figure 3
Let L1 and L2 be given by the following
1
c' d' e'
SL2
Figure e
0
Figure 4
Let S1 = Sc, let S2 = S and let ce = x (thereby deter
1 c' X x'
mining 9). Apply Theorem 2.1.22 to L1, L2, Sl, S2, and 8
(noting that all of the hypotheses are satisfied) to obtain
the following orthomodular lattice (for simplicity of
notation, we write z for [z]).
1
a' b'# c d# e.
Lo
0
Figure 5
This lattice will be used in the following proposition.
2.1.26 Proposition. Let the hypotheses of Theorem
2.1.22 be satisfied, then the lattice L of the conclusion
of Theorem 2.1.22 is complete. Moreover, if L1 and L2 (of
that theorem) are atomic, then L is atomic and the atoms of
L are of the form [m] where m is an atom of either L1 or
L2; however, not every element of L of the form [x], where
x is an atom of L1 or L2, need be an atom of L.
Proof. To show that L is complete, let M be any
subset of L. Let N = ([x]EM:there exists x E[x]], let
N1 = (x:xEL1 and [x]EN), let n = sup1 N1 (the supremum
exists since L1 is complete), let P = MN, let
P2 = [x:xEL2, [x]EM, and [x] = [x2}), and let p = sup2 P2
(the supremum exists since L2 is complete).
We claim that sup N exists and equals [n]. It is
clear that [n] a [x] for all [x]EN. Assume that [z] a [x]
for all [x]EN. If there exists zlE[z], then z1 ~1 xl for
all xEN1; hence, in this case, z1 a1 n so that [z] a [n].
If [z] = (22), then for each xEN1 there exists
[w]x = (w,w2] such that z2 a2 w and wx a1 x. Hence
wx a1 n so that [w] 2 [n]. Consequently sup N exists and
equals [n].
We now claim that sup P exists and equals [p].
It is clear that [p] a [x] for all [x]EP. Assume that
[z] a [x] for all [x]EP. If there exists z2E[z], then
z2 !2 x for all xEP2; hence, in this case, z2 a2 p so that
[z] a [p]. If [z] = (zl), then, for each xGP2, there
xx x x
exists [w]x = [W1,W2 such that z1 a w1 and w2 a2 x.
Hence wx a2 p so that [z] a [p). Consequently sup P exists
and equals [p].
Since L is a lattice [n]v[p] exists in L. We
claim that [n]v[p] = sup M. Clearly [n]v[p] a [x] for all
xEM. If [z] a [x] for all [x]EM, then [z] a [n], [z] a [p],
and therefore [z] a [n]v[p]. Consequently sup M exists and
equals [n]v[p]. Since L is orthocomplemented, inf M also
exists. Therefore L is complete.
To show that L is atomic provided L1 and L2 are
atomic, let [x] be any nonzero element of L. Then we may
assume there exists a representative, xl, of [x] from L1.
Since L1 is atomic, there exists an atom a EL1 such that
al i x1. Either [al] is an atom of L or there exists
[y]EL such that [0] < [y] < [all. Assume the latter, then
[y] = [y2); and since L2 is atomic, it follows that y2
dominates some atom b2 of L2. Then
[0] < [b2] [y] < [al] [x]. Now [b2] is an atom of L,
for, if it were not then either al or b2 would fail to be
an atom in L1 or L2, respectively. Hence every nonzero
element of L dominates some atom [m] of L where m is either
an atom of L1 or an atom of L2.
To show that not every element of L of the form
[x], where x is an atom of L1 or L2, need be an atom of L
consider the following lattices:
Figure 6
Let L1 be the (Boolean) lattice given in Figure 6,
let S1 = (0,1,p,p',k,k',h,h'}, let L2 be the (orthomodular)
lattice given in Figure 5, let S2 = (0,l,a,a',b,b',c,c'),
and let
9 = {(0,0),(1,1), (a,p'),(b',k'), (c',h') ,(a',p),(b,k) ,(c,h)].
Apply Theorem 2.1.22 to obtain the lattice given in Figure
7. (Change the notation by dropping the brackets and
representing all twoelement equivalence classes by the
representative in L1.)
1
Figure 7
Now [p'] = [a] is not an atom of L but a is an atom of L2.
Hence not every element of L of the form [x], where x is an
atom of L1 or L2, need be an atom of L.
2. A Partial Converse
2.2.1 Remark. We have shown that if an equivalence
relation is defined on the union of two disjoint ortho
modular lattices in such a way that elements of isomorphic
sections (of a certain type) are equivalent to one another,
then, upon "dividing out" this equivalence relation, a new
orthomodular lattice is obtained. (We pictorially think of
the elements of one section as being "pasted" to the
corresponding elements of the other section.)
We now prove a partial converse of Theorem 2.1.22.
We maintain Convention 2.1.1 and make the further
assumption that L (cf. Definition 2.1.4) is an orthomodular
lattice. After obtaining some preliminary more general
information, we make the additional assumption that L1 and
L2 are complete Boolean lattices.
The preliminary information necessary (cf. Lemma
2.2.2, Corollary 2.2.3, and Lemma 2.2.4) is concerned with
comparabilities which may not obtain and restrictions on
those that may. These observations form the crux of our
main arguments.
2.2.2 Lemma. Assume that L is an orthomodular lattice
and that Si < Li. Let [z],[x],[y],[u]EL be such that
[z] s [x],[y] and [x],[y] a [u].
(1) If [x] = (x1) and [y] = [y2], then [u]Es.
(2) If [x] = [x2) and [y] = (yl}, then [u]ES.
(3) If [x] = fxli and [y] = {y2}, then [z]ES.
(4) If [x] = [x2) and [y] = [yl), then [z]ES.
Proof. (2) follows from (1) by symmetry; (3) and
(4) are dual to (1) and (2), respectively. Hence we need
only prove (1).
Suppose that (1) is false. Then either [u] = {(ul
or [u] = (u21 obtains. We may assume that [u] = 1ul}. Now,
u1 ! X1' and there exists [w]ES such that ul 1~ w1 and
w2 2 Y2. Let [a] = [u]A[y]', then [a]' = [y]v[u]'. Since
L is an orthomodular lattice and
(I) [y] 5 [w] < [u],
by the OMI and DOMI we have
(II) [u] = [y]v([u]A[y]') = [y]v[a]
and
(III) [y] = [u]A([y]v[u]') = [u]A[a]'.
Moreover, because of (I), by taking the infimum of both
sides of (III) with [w], we obtain
[y] = [w]A[y] = [w]A([u]A[a]')
= ([w]A[u])A[a]' = [w]A[a]',
i.e.,
(IV) [y] = [w]A[a]'.
,
If there exists alE[a], then, since Si < Li (cf. Lemmata
2.1.15 and 2.1.16), by (IV) we have
[Y2] = [l]A[al] = [w1 A1 al]. Hence [y]ES, contradicting
the fact that [y] = {y2). If there exists a2E[a], then,
since S. < Li (cf. Lemmata 2.1.15 and 2.1.16), by (II) we
have [u1] = [Y2]v[a2] = [Y2 V2 a2]. Hence [u]ES
contradicting the fact that [u] = f(ul.
2.2.3 Corollary. Assume that L is an orthomodular
lattice and that Si < Li. If x EL S1 and Y2EL2S2, then
[xl]v[y2]ES. If x2EL2S2 and yfEL1S1, then [x2]v[Y1]ES.
Proof. To prove the first statement let
[x1]v[y2] = [u] in Lemma 2.2.2. The second statement
follows similarly.
2.2.4 Lemma. Assume that L is an orthomodular lattice
and that Si < Li. Then there do not exist [a],[b]EL such
that both [a] s [b] and either
(1) [a] = (al), and [b] = {b2), or
(2) [a] = (a2], and [b] = (b1 .
Proof. By symmetry we need only prove (1).
Suppose the statement is false, then there exists [x]EL
such that [x]:[a] & [b], [a] = [al}, and [b] = fb2]. Then
[x]ES, and, since [a] t [x] s [b], by the DOMI we have
[a] = [b]A([a]v[b]'). Hence
[a] = [x]A[a] = [x]A([b]A([a]v[b]')) = ([x]A[b])A([a]v[b]')
= [x]A([a]v[b]'),
and therefore
[a] = [x]A([a]v[b]').
But by Corollary 2.2.3 [a]v[b]'ES and [x]ES. Hence [a]ES,
which is a contradiction. Therefore no such elements
exist.
Thus ends our preliminary results. We are now
prepared to deal with Boolean lattices. Henceforth, in
addition to Convention 2.1.1 and the assumption that L is
an orthomodular lattice, we assume that L1 and L2 are
complete Boolean lattices. In each of the following
results all assumptions augmenting Convention 2.1.1 will be
made explicit. With the exception of direct reference to
the lattices L1 and L2 of Convention 2.1.1, we will denote
a Boolean lattice by the symbol B.
2.2.5 Lemma. Let B be a Boolean lattice, and let T be
a proper suborthomodular lattice of B. If zEB is such
that B(0,z) c T, then there exists mEB(z,l) such that mT.
Proof. Suppose the statement is false. Then
there exists zEB such that B(O,z) c T and B(z,l) c T. Then
B(0,z') c T since T is a suborthomodular lattice of B.
Let xEB. Then x = (xAz)V(xAz') and consequently xET since
xAzET and xAz'ET. Therefore B c T, i.e., B = T, which
contradicts the fact that T is a proper sublattice of B.
2.2.6 Lemma. Let B be a complete Boolean lattice. Let
T be a subcomplete suborthomodular lattice of B. Let
M = (xEB:B(O,x) c T). Let c = sup M. Then cEM and, for
all xEB, B(O,x) c T if and only if x i c.
Proof. To prove that cEM, it suffices to show
that B(0,c) c T. Let a s c. Then
a = aAc = aA(sup M) = sup (aAm:mEM}. But, for all mEM,
(aAm)ET since aAm m and B(0,m) c T. Moreover, since T is
a subcomplete suborthomodular lattice of B, aET.
Consequently B(O,c) c T.
Now if xEB is such that B(O,x) c T, then x6M.
Hence x c. Moreover, if xEB is such that x 9 c, then
B(O,x) c B(O,c) c T.
2.2.7 Lemma. Let each Li of Convention 2.1.1 be a
complete Boolean lattice, let each Si be a subcomplete sub
orthomodular lattice of L., let M. = {xEL.:L.(0,x) c Si),
and let c. = sup M. (i = 1, 2). Then
(1) Li(O,ci) c Si, i.e., ciEMi,
(2) Li(0,xi) c Si if and only if xi gi ci'
(3) xEM1 if and only if xe'M2, and
(4) c19 = c2.
Proof. (1) and (2) are immediate consequences of
Lemma 2.2.6.
To prove that if xEM1 then x6EM2, assume that the
statement is false. Then there exist xEM1 and mEL2(Ox9)S2.
Also, by Lemma 2.2.5 there exists nL1IS1 such that
x < n < 1. Then [m] < [x] < [n] which contradicts Lemma
2.2.4 since mEL2S2, xES1, and nEL1S1. Hence if xEM1,
then xE9M2. A similar argument utilizing 81 in place of 8
proves the converse. Hence (3) is proved.
Now by (3) and the fact that 9 is a complete join
homomorphism (cf. Lemma 1.3.2)
c1a = sup (m9:mEM1l = sup (m:mEM2) = c2. Hence (4) is
proved.
2.2.8 Lemma. Let each Li of Convention 2.1.1 be a
complete Boolean lattice, let each S. be a subcomplete
suborthomodular lattice of Li, let
Mi = (xELi:Li(0,x) c Si], and let ci = sup M (i = 1, 2).
Then the following statements are valid.
(1) If xES S #, then there exists aEL1S1 such
1
that a <1 x.
(2) If xESiSc#, then there exists bELLS1 such
that x
(3) If xES2S +, then there exists aEL2S2 such
2
that a <2 x.
(4) If xES2S +, then there exists bEL2S2 such
2
that x <2 b.
Proof. Ad (1). Let xESISc#, then x ; c. Hence
1
by Lemma 2.2.7 part (2) L(0,x) ; SL, i.e., there exists
aEL S1 such that a <1 x.
Ad (2). Let xES1S # and suppose that there does
1
not exist bEL1S1 such that x <1 b. Then L(x,l) c S1 so
that L(0,x ) c S1. Hence by Lemma 2.2.7 part (2) x cl
and therefore xES #, which is a contradiction. The result
c1
follows.
Ad (3) and (4). These are proved by noting the
symmetry of the hypotheses.
2.2.9 Proposition. Let each Li of Convention 2.1.1 be
a complete Boolean lattice, let each Si be a subcomplete
suborthomodular lattice of Li, let Mi = xELi:Li(0,x) CSi],
and let ci =sup Mi (i = 1, 2). Then S1 = Sc # and S2 = S +.
1 2
Proof. By the symmetry of the hypotheses we
need only prove that S1 = Sc#. By definition of cl,
1 1
is false. Then there exists xES1Sc#. By Lemma 2.2.8
1
part (1) there exists aEL1S1 such that a <1 x. Moreover,
x8ES2S +, and by Lemma 2.2.8 part (4) there exists bEL2S2
2
such that x8 <2 b. Then [a] < [x] < [b], aEL S1, xES1,
and bEL2S2 which contradicts Lemma 2.2.4.
The following theorem is now immediate.
2.2.10 Theorem. Let (L1,c1, ) and (L2,2,+ ) be two
disjoint complete Boolean lattices each of cardinal number
strictly larger than two. Let S. < L. (i = 1, 2) be
subcomplete suborthomodular lattices of L such that there
exists an orthoisomorphism 8:S S2. Let L0 = L1UL2, let
1
P1 = ((x,y)EL0XL0:y = x9), let P = AUPlUP. (By
Proposition 2.1.3 P is an equivalence relation on L0.)
Let L = LO/P. Then L is an orthomodular lattice if and
only if there exists clEL1 and c2EL2 such that S1 = Sc# and
1
S2 = Sc+
2
Proof. Assume that L is an orthomodular lattice.
Let M = (xELi:Li(0,x) c Si}, and let ci = sup Mi
(i = 1, 2). Then by Proposition 2.2.9 S1 = Sc# and
1
S2 = S +. Conversely, if S1 = S # and S2 = S +, then by
1 2
Theorem 2.1.22 L is an orthomodular lattice.
3. Variation on a Theme
In case S1 and S2 consist of only two elements
(0,1), L is said to be the horizontal sum (disjoint sum) of
L and L2, and L is written HS(L1,L2). The simplicity of
the operation HS allows us to make a more general
definition.
Definition. Let (L ,< ), aEI, be a family of
2.3.1
orthomodular lattices of cardinality larger than 2, indexed
by the set I, where I > 1, such that if a and B are
distinct elements of I, then LaniL = Let L0 = U L
cEI
let 1 and 0 denote the unit and zero, respectively, of
L Define a relation P1 on L0 by
P1 = ((x,y)EL XL0:there exist a,BEI such that x = 1a and
y = 1 or x = 0a and y = 0Og}U
where A = ((x,x):xELO). P, is clearly an equivalence
relation. Let L = LO/P. Then L is called the horizontal
sum (disjoint sum) of the lattices L written
L = HS(L :aEI).
2.3.2 Remark. If [x],[y]EL, then we write [x] a [y] in
case there exist representatives xE[x], yE[y], and aEI such
that x,yELa and x 9a y, where :a is the partial ordering in
L .
is easily seen to be a partial order for L;
moreover, for any aEI, [0 ] and [1 ] are the zero and unit,
respectively, for L. Since [0 ] and [1 ] are independent
of a, we write [0] for [0 ], [1] for [a ].
The orthocomplementations in La induce an ortho
complementation in L defined as follows:
for [xa]EL, [x ]' = [x]
where x. is the orthocomplement of xa in La. Since, for
0 ,1 EL 0' = 1 and 1' = 0 and since all other
a a a a a a'
equivalence classes are singletons, ':L > L is a well
defined orthocomplementation for L.
2.3.3 Lemma. Let L = HS(L :aEI). Then L is an ortho
modular lattice.
Proof. Let [x],[y]EL. Then an immediate conse
quence of the definition of A is the fact that [x]v[y]
exists and
[x V y] if there exists aEI such that x,yEL .
[x]v[y] = [1] if there exists a,BEI such that a # 8,
xEL and yELB.
Since L is an orthocomplemented poset such that the
supremum of every pair of elements exists in L, it follows
that L is an orthocomplemented lattice. To prove that L is
an orthomodular lattice, let [x] s [y]. If [x] = [O],
[x] = [1], [y] = [0], or [y] = [1], then the computation of
the OMI is trivial. If none of these hold, then [x] and
[y] are singletons whose only representatives are x and y,
respectively, and there exists aEI such that x,yEL Hence
[x]V([y]A[x]') = [x]v([y]A[x']) = [x]v[y Aa X']
= [x V (y A x')] = [y]. Therefore L is an
orthomodular lattice.
2.3.4 Definition. Let P be an orthocomplemented poset.
Then a section S of P is said to be simple if and only if
S(0,1l consists only of atoms and coatoms of P.
The following theorem is a variation of Theorem
2.1.22. 1he function 6 is no longer required to be an
order homomorphism, and additional assumptions are made on
L1, L2, S1, and S2'
2.3.5 Theorem. Let Si be a simple section of the
orthomodular lattice Li (i = 1, 2), and let :SS1 2 be an
orthoisomorphism. Assume that L1 = HS(L :aEI), for some
index set I, and that, for any aEI, S nL has cardinal
number 2 or 4. Moreover, assume that 06 = 0 and that xES1
and x an atom of L1 imply xG is an atom of L2. Define P,
R, L, and as in Definitions 2.1.2, 2.1.4, and 2.1.11.
Then L is an orthomodular lattice.
Proof. The proof that P is an equivalence
relation is exactly the same as in Proposition 2.1.3.
However, since 8 is not an order isomorphism, the proof
that R is a partial order is not the same as Proposition
2.1.9, but it follows the latter proof closely enough
that we shall, in general, only point out those parts which
are different. (In this proof the symbols (I), (II),
(III), and (IV) refer to the comparabilities bearing these
labels and appearing in the proof of Proposition 2.1.9;
moreover, all unidentified elements are defined in the
corresponding parts of the proof of Proposition 2.1.9.)
(1) R is reflexive since [x]:[x]R[x].
(2) To prove that R is antisymmetric, assume
that [z]:[x]R[y] and [w]:[y]R[x]. We must show that
[x] = [y]. There are nine cases as listed in Proposition
2.1.9. We may assume [[x],[y]]n{[0],[l]} = b.
Case (1). [x] = (xl) and [y] = (yl). Since S1
is a simple section (I) obtains.
Case (2). [x] = (xl1 and [y] = (y2]. [z] must
be an atom or a coatom of L. Each possibility is contra
dictory.
Case (3). [x] = {xl) and [y] = (y1,Y2).
implies [y] is a coatom and [y]R[x] implies [y] is
atom. Hence [x] = [y] (which in this case is a
contradiction).
Case (4). [x] = (x2) and [y] = tyll. As
Case (2) this is contradictory.
Case (5). [x] = (x2] and [y] = (y2). Sin
is a simple section (II) obtains.
Cases (6), (7), and (8) are resolved as in
[x]R[y]
an
in
ce S2
Case (3).
Case (9). [x] = (Xl,x21 and [y] = {ylY2l.
Suppose [x] # [y]. Then [x]R[y] implies xl is an atom but
not a coatom of L1 and yl is a coatom but not an atom of
L1; [y]R[x] implies y, is an atom but not a coatom of L1.
This is a contradiction.
(3) To prove that R is transitive we must again
consider the sixteen cases provided by (IV). Cases (1),
(3), (4), and (5) are proved exactly as before. Case (2)
is resolved as follows: If [x] = [0], then[r]:[x]R2[y]. If
[x] = [y], then [x]:[x]R [y]. If [y] = [1], then
[p]:[x]R2[y]. Hence we may assume [0] t [x], [x] # [y],
and [y] # [1]. Then [x] = [p] or [p] = [y]. In the former
case [p] = [q] implies [q]:[x]R2[y], and [p] # [q] implies
[q] = [r] = [y] and [p]:[x]R R2[y]. In the latter case
[p] = [q] = [r] = [y] and [p]:[x]RR2[y].
In Case (6), if [x] = [0], [x] = [y], or
[y] = [1], then [r]:[x]R1R2[y]. Hence we may assume
[x] # [0], [x] X [y], and [y] ? [1]. Since [p] = {plP2,
[q] = (qlq2), [r] = (rl,r2), xl1 pi' P2 "2 q2' q1 '1 rl
and r2 !2 y2, it follows that [p], [q], and [r] are atoms
or coatoms. Hence either [p] = [q] or [q] = [r]. If
[p3 = [q] and [p] is an atom, then [p] = [x] and
[r]:[x]R1R21[]; if [p] = [q] and [p] is a coatom, then
[p] = [y] and [p]:[x]R1R2[y]. Hence we may assume
[p)] [q]. It follows that [x] = [p] and [q] = [r] = [y]
and therefore [q]:[x]R R2[y]. Consequently, in all cases,
([x],[y])ER1R2.
In Case (7) we may assume [x] X [0], [x] A [y],
and [y] A [1]. If [p] = [r], then [p]:[x]R2[y]. If
[p] [r], then [x] = [p], [r] = [y], and [q]:[x]R2[y].
Hence ([x],[y])ER2UR2 C R.
In Case (8) we may assume [x] 4 [0], [x] $ [y],
and [y] 4 [1]. If [p] = [r], then [p]:[x]R2[y]. If
[p] [r], then [x] = [p], [r] = [y], and [q]:[x]R2R1[y].
Since we have not made use of the hypotheses that
L = HS(L:a~EI) and that, for any aEI, (S nLa) = 2 or 4,
we may obtain Cases (9) (16) by symmetry of hypotheses.
Consequently L is a poset.
We must now prove that ':L > L is a well
defined orthocomplementation. That it is welldefined is
proved exactly as in Proposition 2.1.14.
Assume that [x],[x]' 9 [a]. We must prove that
[a] = [1]. Suppose that [a] # [1]. Then [x] # [0],[1].
If [a] = (al), then, since the only twoelement equivalence
classes are [0], [1], and [m] where [m] is either an atom
or a coatom of L, either [x] = (xl or [x] = (x1,x2). If
[x] = (x1], then we have xl,x 9l al and hence
[a ] = [a] = [1]. If [x] = {xlx2}, then x, is either zero
or an atom of LL; it cannot be zero since [x]' & [a], hence
xI is an atom of L1 and therefore x1 is a coatom of L1
which contradicts the fact that [x]' < [a] = (a1}. If
[a] = (a2), a similar argument yields a contradiction. If
[a] = {al,a2], then al is either an atom or a coatom of
L1. It cannot be an atom since [x],[x]' [a]. Hence al
is a coatom of L1. If [x] = (xl), then xl,x' 1 al and
hence [a] = [1]. If [x] = (x2}, then there exist [w],[z]
such that x2 12 w2 wl al, x2 2 2', and z 1I a Now
[w] and [z] must be coatoms of L, hence w1 = zI = al and
w2 = z2. Therefore x2,x+ 2 z2 and hence [z] = [a] = [1].
If [x] = (xl,x2], then [x] is an atom of L, [a] is a
coatom of L, and [x]' s [a] yields a contradiction. Hence
[a] = [11, and consequently [x]v[x]' exists and equals [1].
Dually [x]A[x]' exists and equals [0]. Therefore ':L > L
is an orthocomplementation.
To show that L is a lattice we need only show
that [x]v[y] exists for every [x],[y]EL. Hence assume
[x],[y]EL. We need only consider the following six cases:
Case (1) [x] = [(xl and [y] = {y1),
Case (2) [x] = (xl) and [y] = {y2],
Case (3) [x] = (xl] and [y] = (ylY2,
Case (4) [x] = (x2] and [y] = [y2],
Case (5) [x] = (x2) and [y] = (yl'Y2]'
Case (6) [x] = (xl,x2) and [y] = (yly2l.
Ad (1). [x1 V1 y1] a [x],[y]. If [z] z [x],[y],
then there exists z E[z] such that z 1 1 xI v, y1 and hence
[z] [x1 V1 y]. For, if [z] = (z2], then there exists
[w] = {wl,w2) such that z2 "2 w2 and w1 1 x1; in this case
[w] is a coatom of L or [1] which yields a contradiction
since [z] = {z2).
Ad (2). Let [z]EU([x],[y]). It follows that
[z] = [1] or [z] = (zl,z2) is a coatom of L. We may
assume [z] is a coatom of L. Since, by hypothesis,
L1 = HS(L :aEI), xiEL$ for some BEI; it follows that
#(S1nL) = 4, zlESInL8 and [z] is unique with these
properties. Consequently [z] = [x]v[y].
Ad (3). Note that [xl v1 y1] a [x],[y]. Let
[z] a [x],[y]. If [z] = (zl, then [y] is an atom of L,
z1 a1 Y1, and z1 l xl hence zl ,1 x1 V1 y1 and
[z] a [x1 v1 y ]. The case [z] = {z2] is contradictory
since [z] 2 [x] and Si is a simple section of Li
(i = 1, 2). If [z] = (z1,z21, then [z] is a coatom of L,
z1 2 X, and z1 21 y1; hence [z] a [xl v1 yl].
Ad (4). The proof, in this case, is similar to
that of (1).
Ad (5). The proof, in this case, is similar to
that of, (3).
Ad (6). We may assume that
([x],[y]]n([[],[1]j = 0 and that [x] / [y]. We claim that
[x]v[y] = [x2 V2 y]. Clearly [x V2 y2 ] a [x],[y]. Let
[z] a [x],[y]. We may assume that neither [x] nor [y] is a
coatom in L, and hence that both [x] and [y] are atoms in
L. [z] $ ([zl, since [z] = (zl] contradicts the fact that
#(LanS1) 1 4, for every aEI, since xl,Y1 are (in this case)
atoms of the same L If [z] = [z2], then z2 22 x2'Y2 so
that z2 a2 x2 v2 y2 and hence [z] a [x2 V2 y2]. If
[z] = [zL,z2], then [z] = 1 or [z] is a coatom of L. We
may assume that [z] / 1; hence [z] is a coatom of L. The
time has come to note that x v1 yl = 1 since #(S1L ) a 4
for each aEI. Hence z2 22 x2 and z2 2 Y2; therefore
z2 2 x2 v2 2 and [z] a [x2 v2 y2]. Hence in this case
[x]v[y] = [x2 v2 Y2]. Consequently L is a lattice.
To show that L satisfies the OMI, assume that
[x] s [y]. We must show that [y] = [x]v([y]A[x]'). We may
assume that [0] < [x] < [y] < [1]. We consider nine cases.
Case (1). [x] = xl), and [y] = {yl). Since S1
is simple, all computations occur in L1 which is ortho
modular. Hence the OMI is satisfied.
Case (2). [x] = (xlj, and [y] = [y2). Then
there exists [z] = fzl,z2) such that xl ~1 z1 and z2 S2 Y2
But this contradicts the fact that S. is a simple section
(i = 1, 2).
Case (3). [x] = (xl), and [y] = {yl1Y2]. Then
[y] is a coatom of L. It follows that xl 1 y1 and hence
that the OMI is satisfied since all computations are made
in L1.
Case (4). [x] = (x2), and [y] = (yl). The proof
is similar to that of Case (2).
Case (5). [x] = {x2}, and [y] = (y2l. The proof
is similar to that of Case (1).
Case (6). [x] = (x2), and [y] = (yl,Y2]. The
proof is similar to that of Case (3).
Case (7). [x] = {x1X2]), and [y] = (yl). The
proof is similar to that of Case (3).
Case (8). [x] = (x ,x2), and [y] = [y2]. The
proof is similar to that of Case (3).
Case (9). [x] = fxl,x2), and [y] = (yL,y2}. Then
[x] is an atom of L, [y] is a coatom of L, and [y] / [x]'.
It follows that x2 2 y2 since L1 = HS(L :aEI) and
(SlnL ) 4 for each aEI. Therefore all computations are
made in L2 and hence the OMI is satisfied.
Consequently L is an orthomodular lattice.
2.3.6 Remark. Let L = HS(L :aEI). If at least one La
admits a chain of length 3, then L is nonmodular.
Proof. Assume that L admits a chain of length 3.
Then there exist a,bEL (0,1l such that a < b. Let
cELO(0,1l for any 8 # a. Then M(c,b) does not hold.
Hence L is not modular.
The only (previously) known finite orthomodular
nonmodular lattices are those horizontal sums admitting
at least one chain of length 3 and the following lattice
(see Figure 10) given by R. P. Dilworth in [1]. Rather
than exhibit this lattice "out of the blue" we construct it
by applying Theorem 2.3.5.
2.3.7 Example. Let L1 and L2 be the following
orthomodular lattices:
1
a' b' cg e' 'f g'
L =
a b c e f g
0
Figure 8
1
x' d' y
L2
x d y
Figure 9
Let S1 = [0,l,c,c',e,e'5 and S2 = (0,l,x,x',y,y']. Define
e:S I S2 by 9 = ((0,0),(1,1),(c,x),(c',x'),(e,y),(e',y')}.
Note that the hypotheses of Theorem 2.3.5 are satisfied.
Hence the lattice given in Figure 10, L, is an orthomodular
lattice. Relabeling the equivalence classes {a), (b},
(c,x), (d], (e,y), {f}, and (g) by a, b, c, d, e, f, and g,
respectively, we obtain the following diagram for L:
Figure 10
L is frequently denoted by the symbol D16.
nonmodular since M(g,a') fails to hold.
Note that D16 may be constructed
applications of Theorem 2.1.22. The first
Theorem 2.1.22 yields the lattice given in
Note that L is
from 23 by two
application of
Figure 5. Form
Figure 11
Let L1 be the lattice given in Figure 5, let
S1 = (0,1,e,e'}, let L2 be the lattice given in Figure 11,
let S2 = (0,l,z,z'], and let
8 = ((0,0),(1,1),(e,z),(e',z')]. Apply Theorem 2.1.22 and
obtain D16 (given in Figure 10) by dropping the brackets
and representing twoelement equivalence classes by the
representative in L1.
In [3], the process of applying Theorem 2.1.22
with L2 = 23 and with simple fourelement sections was
baptized adiunction of a crown to the lattice L We now
prove a theorem which generalizes the method of adjoining
crowns and which allows us to exhibit a countable family
of nonisomorphic finite orthomodular nonmodular lattices
The construction is a generalization of the second method,
exhibited above, of constructing D16.
2.3.8 Theorem. For each i = 1, 2, 3, let L. be a
complete orthomodular lattice admitting a chain of length
3, let xiELi be such that the principal section Sx is a
1
proper suborthomodular lattice of Li, and let ylEL, be
such that S ylnn = (0,1. Let there exist a,bEL Sxi
y1 x1 i xi
i = 2 or 3, such that a < b. Let S be orthoisomorphic
to Sx and let Sy be orthoisomorphic to Sx. Let L be
x 2 1 x3
the orthomodular lattice obtained by applying Theorem
2.1.22 to L1, L2, S l, and Sx2; then the principal section
1 2 q ^
S = [[z]EL:zES y is a proper suborthomodular lattice
(which is orthoisomorphic to S ) of L. Let L be the
orthomodular lattice obtained by applying Theorem 2.1.22
to L, L3, Sy1, and S3. Then L is nonmodular.
Proof. S is orthoisomorphic to S For, the
function x > [x] is clearly an orthoisomorphism. To
show that it is an epimorphism, assume [z]ES ; then we may
assume that [yl] a [z]. Since S ylnS = [0,1),
[yl] = fyl). We claim that there exists zlE[z]. For, if
[z] = (z2), then there exists [w] = fwl,w2} such that
Yl '1 Wl and w2 s2 z2. Then w ESy lSxl. Hence w1 = 1 and
therefore z2 = 1 which contradicts the fact that
[z] = [z2]. Consequently there exists zlE[z] and y al zl;
therefore zlESyI and z1 > [z]. Hence x > [x] is an
epimorphism.
There exist a,bEL2Sx2 such that a <2 b, and, in
L, [a] = (a) and [b] = [b}; moreover, in L, [[a]] = ((a)]
and [[b]] = [[b}3. Now there exists cEL3S3
[[c]] = ({c)). Let [[m]] = [[b]]A[[c]]. By the dual of
Corollary 2.1.20, [[m]]ES. Hence [[m]] has a
representative [m]S y. It follows that [m] e [yl]'
Hence if [m] 4 0, then since S nS = (0,1), [m] is a
y1 xl
69
singleton. But [c] is a singleton. Hence by Lemma 2.1.19,
[m] ; [c]. This is a contradiction. Hence [m] = [0].
Similarly, [[a]]v[[c]] = 1. Now [[a]] < [[b]] and
[[a]]v([[c]]A[[b]]) = [[a]]VO < 1A[[b]]
= ([[a]]v[[c]])A[[b]].
Hence M([[c]],[[b]]) does not obtain and consequently L is
nonmodular.
CHAPTER 3
THE REPRESENTATION OF ORTHOCOMPLEMENTED
POSETS BY SETS
1. Fsequences
In [7] Zierler and Schlessinger prove that every
orthocomplemented poset P may be (in some sense) "embedded"
in a Boolean lattice of sets. The set utilized is the set
of all order orthohomomorphisms of P into the twoelement
Boolean lattice. The proof is made by noting that the
kernel of such homomorphisms is a maximal ideal (of a
special type) and by developing some of the properties of
these ideals.
In the sequel we give a proof for this theorem
which is valid in orthocomplemented posets which are, in
some sense, "not too wide." In particular, our proof
holds for all finite orthocomplemented posets. The
advantage of our method is that it prescribes an algorithm
for generating the order orthohomomorphisms. Because of
this we are able to associate a certain matrix, called a
deterministic Travis matrix, of zeroes and ones with such
posets. The columns of this matrix, partially ordered
vectorwise, form an orthocomplemented poset which is
orthoisomorphic to the given poset.
70
3.1.1 Convention. Let P be an orthocomplemented poset.
3.1.2 Notation. (1) Let N denote the set of natural
numbers, let R denote the set of real numbers,
and let [0,1] denote the closed unit interval
in R.
(2) Let 2P denote the power set of P.
(3) If X and Y are two sets and if (x,y)EXXY, then by
nl((x,y)) we mean x, and by n2((x,y)) we mean y.
(4) For M c P, let XM denote the characteristic
function of the set M, i.e., yM:P > R is
defined as follows: for xEP,
f1 if xEM.
XMX) O0 if xM.
3.1.3 Definition. Let xEP. Define f :S > (0,1] c
x x
as follows: for zESx, let fx(z) = XL(x,l)().
3.1.4 Remark. Let xEL.
(1) If x = 0, then fx(z) = 1 for all zESx
1 if x z.
(2) If x Z 0, then, for zESx, f (z) = if x
0 if x & z'.
(3) The range of fx is a subset of [0,1].
Proof. These are immediate consequences of the
fact that Sx = L(x,l)UL(0,x'), the definition of fx, and
the fact that P is an orthocomplemented poset.
3.1.5 Definition. A rsequence on P is a function
y:N > PX2 such that the following conditions hold:
(1) If y(i) = (b,B), then b = 0 if and only if B = 0.
(2) If y(i) = (b,B), then B 0 0 implies bEB.
(3) n2(Y(1)) = P.
(4) If y(i) = (0,0), then y(i+l) = (0,0).
(5) If y(i) = (b,B), b # 0, then y(i+l) = (c,C)
where C = BSb.
If y is a rsequence, then define
B = [7l(Y(i)):iE6N([)
and
D = U S
Y xEB x
Y
3.1.6 Notation. In the sequel y always denotes a
rsequence. If only one y is under discussion, then we
sometimes write bi for Trl(Y(i)) and Bi for r2(y(i)); then
y(i) = (bi,Bi).
3.1.7 Lemma. Let y be a rsequence. Then the
following statements are valid.
n
(1) Bn+1 = P( Sb ).
j=l j
(2) If i < j and Bi 0 0, then Bi Z Bj.
(3) If P < m, then there exists mEN such that
n : m implies B = 0.
(4) If P < , then D = P.
Y
(5) If a,bEB a / b, then aESb or bOSa.
(6) If a,bEB xESanSb, then fa(x) = f (x).
Proof. Ad (1). If n = 1, then
1
B2 = BlSb = L( U Sb ). Assume that the formula is valid
for j=l j
for n = k1. If bk 0, then
ki k
B+l= BkSb = (P(U) Sb ))S = P(U Sb)
k j=l j k j=l
then Bk = 4 and therefore Bk+1 = 0. But
k
P( Sb ) c PS = PP = 6. Hence Bk+l = 0
j=l j bk
If bk = 0,
k
= P(U Sb
j=l j
Ad (2). Since biEB ilSb.
i
B i BSb = B+ B .
1
Ad (3). Suppose the statement is fali
there exists a rsequence y such that Bi ? d foi
S1
and by part (2) P = B1 2 B2 B3 .... Hence
P > #B1 > B2 > 3 > ..., and #B > 0 for a
this is a contradiction since there does not exj
infinite strictly decreasing sequence of natural
less than the natural number (P)+l.
se. Then
r all iEN
ii iEN. I
ist an
i numbers
Ad (4). Suppose the statement is false. Then
there exists xEPD Let m = inf (i:b. = 0). Now m > 1
since B1 = P # 0 (cf. Definition 3.1.5), Bm = and
m1
bm_ i 0. But, by part (1), xELD = L( U S = B = 0
m1y i1 b m
which is a contradiction.
Ad (5). Let a = nl(y(i)) and b = T2(y(j)); then
i 7 j. If i < j, then bEBj c Bi+1 = BiSa; hence b9Sa. If
j < i, then aEBi c Bj+1 = BjSb; hence aOSb.
Ad (6). Suppose the statement is false. Then we
may assume that f (x) = 1 and fb(x) = 0. Hence a s x and
b 9 x'; therefore a 6 x 4 b'. Consequently bESa and aESb
which contradicts part (3).
3.1.8 Definition. (1) For each Psequence y, define
g :D > (0,1) by g (x) = fb(x) whenever bEB
and xESb.
(2) If y,6 are Fsequences, we define y a 6 in case
D = D6 and g (x) = g6(x) for all xED Clearly
a is an equivalence relation. Denote the
equivalence class containing y by [y).
(3) Let Z = ([y]:D = L). Each [y]EZ is said to be a
Fstate; Z is called the Pstate space for P.
3.1.9 Lemma. Let y be a rsequence. Then
(1) g :D > (0,1) is a welldefined function from
D into [0,1].
(2) If xED then x'ED g (x') = 0 if g (x) = 1,
and g (x') = 1 if g (x) = 0.
(3) g (0) = 0 and g (1) = 1.
(4) If xED g (x) = 1, yEP, and x : y, then yEDy
and g (y) = 1.
Proof. Ad (1). By Lemma 3.1.7 part (6), g is
welldefined. The domain is D by definition; the range is
a subset of [0,1] by Remark 3.1.4.
Ad (2). If xED then xESb for some bEB Hence
by Lemma 1.3.6 x'ESb c D Moreover, since b a x if and
only if x' I b', and x a b' if and only if b e x', it
follows that fb(x') = 1 if and only if fb(x) = 0 and
fb(x') = 0 if and only if fb(x) = 1. The result follows.
Ad (3). Since 1ESb for all bEP, it follows that
fb(1) = 1 for all bEP and hence that g (1) = 1. Therefore
by part (2) g (0) = 0.
Ad (4). Since g (x) = 1, there exists bEB such
that 0 < b & x. Hence b i y, and yESb; therefore yED and
g (y) = 1.
Condition I. For every rsequence y, D = P.
Condition II. (1) If aEP(0), then there exists
a Tsequence y such that l1(Y(1)) = a and D = P.
(2) If a,bEP(0}, b'Sa, then there exists a Fsequence
y such that nr(y(l)) = a, nl(y(2)) = b, and
D = P.
Y
3.1.10 Lemma. Condition I implies Condition II.
Proof. Assume that Condition I holds.
(1) Let aEP. Define y(l) = (a,P). For n > 1,
let Fn = T2(Y(nl))Sl(y(nl)) and define
(n) (0,) if F = ~
y(n) n
(f,Fn) for any fEFn if Fn .
Then y is a Fsequence, Tl(Y(1)) = a, and, by Condition I,
D = P.
Y
(2) Let a,bEP be such that bOSa. Define
y(l) = (a,P) and y(2) = (b,PSa). For n > 2, let
Fn = 2(Y(nl))Snl(y(n1)) and define
) (0,0) if F = ~.
y(n) =
(f,Fn) for any fEFn if Fn 0.
Then y is a Psequence, Tnl((l)) = a, nl(y(2)) = b, and, by
Condition I, D = P.
The following examples of horizontal sums show
that the converse of Lemma 3.1.10 is not valid and that
there are posets which satisfy neither.
3.1.11 Example. Let L = HS(24:iEF). A rsequence on L
satisfying (1) (or (2)) of Condition II may be easily
constructed by noting the following fact: for every set of
atoms A of L consisting of exactly one atom of each 2k,
there exists a rsequence y such that B = A (and hence
D = L). But the rsequence defined by
n1
y(n) = (a2n,LU Sa ) where a2n is an atom of 22n has the
i=l 2i
property that
D y 2 2 L.
Y n=1 2n
Hence L satisfies Condition II but not Condition I.
3.1.12 Example. Let L = HS(2a:aEI) where I is any
uncountable index set. Then L satisfies neither Condition
I nor Condition II since, for any rsequence y, D is a
countable union of finite sets and hence countable, whereas
L is uncountable. It follows that for every rsequence y,
C
D p L.
Y
3.1.13 Remark. (1) Condition I implies that [y]EZ for
every rsequence y. Condition II implies that, for every
aEP(O}, there exists [y]EZ such that g (a) = 1; and, for
every a,bEP{0) such that b)Sa, there exists [y]EZ such
that g (a) = g (b) = 1.
(2) By Lemma 3.1.7 part (4), if #P < m, then
Condition I holds. Hence, by Lemma 3.1.10, if P < ,
then Condition II holds.
3.1.14 Lemma. Let P satisfy Condition II, and let
e,fEP. Then e f if and only if g (e) e g (f) for all
[y]EY.
Proof. Assume that e 9 f. We must prove that
g (e) s g (f) for all [y]EZ. Let [y] be any fixed element
of Z, and let yE[y]. We may assume that (e,f}~n{,l1= 0 and
that g (e) = 1. We must show that g (f) = 1. But
1 = g (e) implies that there exists bEB such that fb c g
and fb(e) = 1. Hence b s e so that b 9 f; therefore
fb(f) = 1 and consequently g (f) = 1.
Conversely, assume that g (e) s g (f) for all
[y]EZ. We must prove that e 9 f. Suppose the statement is
false. Then e e f, e ? 0, and f # 1. If f = 0, then by
Lemma 3.1.9, g (f) = 0 for all [y]EZ; hence by hypothesis
g (e) = 0 for all [y]EZ and consequently, by Condition II,
e = 0 which contradicts the fact that e 3 0. Therefore
f f 0. If f r e', then e f' and, by Condition II, there
exists [y]EZ such that g (e) = 1; hence by Lemma 3.1.9
g (f') = 1 and g (f) = 0; consequently 1 = g (e)
which is a contradiction. Therefore f $ e' and
consequently ffs Hence f'kS and by Condition II there
exists [y]EZ such that g (e) = g (f') = 1; therefore by
Lemma 3.1.9 1 = g (e) : g (f) = 0 which is a contradiction.
Consequently e E f.
3.1.15 Theorem. Let P be an orthocomplemented poset
satisfying Condition II, and let e,fEP. Then there exists
a set X and a function I:P > 2X such that
(1) O0 = 0 and 1i = X,
(2) e'5 = X(e$),
(3) e f if and only if eS c f$.
Proof. Let X = Z. Define S as follows:
for eEP, eS = [[y]EZ:g (e) = 11.
(1) follows from Lemma 3.1.9. (2) follows from the fact
that [y]Ee'S if and only if g (e') = 1, and g (e) = 0 if
and only if [y]9et. To prove (3), note that if e a f, then
g (e) = 1 implies g (f) = I for all [y]EZ. Therefore
eS c fS. Conversely, assume that eS c f$ and that [y]ES.
Then g (e) = 1 implies [ly]ce; hence [y]EfS, so that
g (f) = 1. Therefore g (e) = 1 implies g (f) = 1.
Consequently, g (e) < g (f) for all [y]ES, and hence by
Lemma 3.1.14, e t f.
2. Order Orthohomomorphisms
3.2.1 Definition. Let P be an orthocomplemented poset.
Let (0,1) c A, and define 0' = 1 and l' = 0; then (0,L) may
be regarded as an orthocomplemented poset.
(1) Let H = (a:P > [0,l]:(i) e'a = (ea)',
(ii) e < f implies ea s fa, (iii) if
[xi] is a (not necessarily countable)
chain, if inf (xi) exists, and if
x = inf (x.i, then x a = 1 for every
element x. of the chain (xi. implies
xa = 1].
(2) For each aEH, let
E = [xEP:xa = 1 and y < x imply ya = 0}.
(3) For each aEH, let
F = (xEP:there exists a chain ?xi), maximal with
respect to x.a = 1, such that x is the
smallest element of (xi)].
3.2.2 Lemma. If aEH, then Oa = 0 and la = 1.
Proof. It is sufficient to prove Oa = 0.
Suppose the statement is false. Then Oa = 1. Since 0 < 1
implies Oa < la, it follows that la = 1 and hence that
1 = la = 0'a = (Oa)' = 1' = 0 which is a contradiction.
3.2.3 Lemma. Let P be an orthocomplemented poset such
that if (xi) is a chain in P, then inf ([xi exists in P.
Let aEH. Then the following statements obtain.
(1) E j 0 and E = F .
(2) For yEP, ya = 1 if and only if there exists xEEa
such that x : y.
Proof. Ad (1). Let aEH. Then la = 1. (1) is a
chain in P; extend it to a chain (x.i, maximal with respect
to xia = 1. Let x = inf [xi). Since xia = 1, it follows
that xa = 1. By maximality, xE[xi}, therefore x is the
smallest element of (xi.. Since xa = 1, if y < x, it
follows that y = 0; hence xEEa and therefore E f 0.
Clearly Ea 3 F .
Suppose that E a. F Then there exists xEEC
such that for every chain (xi), maximal with respect to
xia = 1, x is not the smallest element of fxi3. Now xEE,
implies xa = 1. Since x is not the smallest element of any
such maximal chain, there exists y < x such that ya = 1.
But xEE and y < x imply ya = 0 which is a contradiction.
Hence E = F
Ad (2). The necessity of the condition is clear.
To prove the sufficiency, note that, since ya = 1, (y] may
be extended to a chain tyi], maximal with respect to
ya = 1. As above, define x = inf {yi). Then xE{yi] and
xEE
3.2.4 Corollary. Let P be an orthocomplemented poset
such that if (xij is a chain in P, then inf [xi. exists in
P. Then for any aEH, P = U S .
xE x
Proof. Suppose the statement is false. Let
yEP(U x), then y'EP( U S ). We may assume ya = 1.
xEE xEE x
ci ci
By Lemma 3.2.3 there exists xEE such that x 6 y. Then
yESx which is a contradiction.
For each Pstate [y], the corresponding function
g :P > (0,1) is an order orthohomomorphism (provided we
regard g as mapping onto (0,1), and define 0' = 1 and
1' = 0). In what follows we prove that every order ortho
homomorphism from a finite orthocomplemented poset into
[0,1) arises in this way.
3.2.5 Proposition. Let P be a finite orthocomplemented
poset, let a be any order orthohomomorphism mapping P onto
(0,1) (regarded as a subset of [0,1] with O' = 1 and
1' = 0). Then there exists a unique Fstate [y] such that
g = a.
Proof. We first note that the third condition on
the elements of H (cf. Definition 3.2.1 part (1)) is
redundant since P is finite; hence H is, in fact, the set
of all order orthohomomorphisms mapping P onto (0,1).
Wellorder Ea; since P is finite, E may be
written as [bl,b2,...,b n for some nER, where bi # bj if
i F j. Define y:R > PX2P by y(i) = (bi,B ) where biEE
if 1 i S n, bi = 0 if i > n, B1 = P, and Bi+1 = BiSb
for all iEN. By the definition of E and the fact that
Oa = 0 for all aEH, it follows that b. # 0 for
1
i = 1, ..., n; hence y is a Psequence. By Corollary 3.2.4,
[y] is a rstate. Finally, by Definition 3.1.8 parts (1)
and (3) and Definition 3.2.1 part (2), it follows that
gy = a and that [y] is unique.
3. Travis Matrices
The following definition is motivated by certain
tables which appear in the 1962 Wayne State University
Master's Thesis of R. D. Travis [6].
3.3.1 Definition. Let 2* and P* be two nonempty sets.
Let ':P* > P* be a mapping of period 2. By a Travis
matrix over (Z*,P*,') we mean a function T:Z*XP* > [0,1]
which satisfies the following postulates. (For simplicity
we write a(e) for T(a,e).)
(1) 0 K a(e) : 1 for all aE.*, eEP*.
(2) a(e') = 1 a(e) for all aE6*, eEP*.
(3) If e,fEP* are such that a(e) = a(f) for all aEZ*,
then e = f.
(4) If a,5E6* are such that a(e) = B(e) for all eEP*,
then a = 8.
(5) There exists OEP such that a(0) = 0 for all aE.*.
(6) k = inf (sup (a(e)) > .
eEP* a.E*
e O
We define the "weak" partial order induced on P* by _* as
follows: Let e,fEP*, then e S* f if and only if
a(e) < a(f) for all aES*.
If the range of T is (0,1), then T is said to be
a deterministic Travis matrix over (Z*,P*,').
3.3.2 Lemma. (P*,s*,') is an orthocomplemented poset.
Proof. Let e,f,gEP*. Since a(e) = a(e) for all
aSE*, e s* e. If e s* f and f <* e, then a(e) s a(f) and
a(f) i a(e) for all acE*; hence a(e) = a(f) for all aEZ*
and therefore e = f by Postulate (3) of Definition 3.3.1.
Moreover, if e S* f and f * g, then a(e) S a(f) S a(g) for
all aE2*; hence e t* g. Consequently 5* partially orders
P*. Since a((e')') = 1 [1 a(e)] = a(e) for all aEZ*,
e' = e. If e S* f, then a(e) 9 a(f) for all aES* and
1 a(f) i 1 a(e) for all aE6*, i.e., a(f') s l(e') for
all aEZ*; hence f' s* e'.
By Postulate (5) there exists E0P. Define 1 to
be 0'; 0 S* e' for all eEP implies that e S* 1 for all eEP.
We claim that eve' exists and equals 1. Suppose e,e' S f;
we must prove that f = 1. Suppose that f # 1. Then f' # 0
and 1 < k a sup (a(f'):aE2*}. Therefore there exists aEZ*
such that < a(f') = 1 a(f); hence a(f) < a. Since
(I) a(e) s a(f) < a,
it follows that
(II) 1 a(e) a(f) < .
By adding the extremes of the inequalities (I) and (II) we
obtain 1 < 1 which is a contradiction. Therefore f = 1.
Similarly, eAe' exists and eAe' = 0. Conse
quently (P*,t*,') is an orthocomplemented poset.
3.3.3 Definition. Let T denote the matrix
T:ZXP > 0,1) defined by T([y],e) = g (e).
3.3.4 Lemma. T is welldefined.
Proof. If [y],[6]EZ and [y] = [6], then D = D6
and g (e) = g8(e) for all eEP.
3.3.5 ,Theorem. If P satisfies Condition II, then T is
Travis matrix over (Z,P,').
Proof. We must show that the following state
ments are valid.
(1) 0 1 g (e) 1 for all [y]EZ, eEP.
(2) g (e') = 1 g (e) for all [y]EZ, eEP.
(3) If e,fEP are such that g (e) = g (f) for all
[y]EZ, then e = f.
(4) If [y],[6]EZ are such that g (e) = g6(e) for all
eEP, then [y] = [6].
(5) There exists 0EP such that g (0) = 0 for all
[y]EZ.
(6) k = inf (sup g (e)) > .
eEP
eEo [y]EY
Ad (1). Since [y]EZ, the domain of g is P. The
inequalities follow immediately from the fact that, for any
bEB eEP, fb(e) equals either 0 or 1.
Ad (2). Let eEP and [y]EZ. Then the result
follows immediately from Lemma 3.1.9.
Ad (3). By Lemma 3.1.14 g (e) g (f) for all
[y]EZ implies that e z f, and g (f) s 9 (e) for all [y]EY
implies that f & e. Hence e = f.
Ad (4). If [y],[6]EE, then D = D = P. There
fore, since g (e) = g,(e) for all eED [y] = [6] by
Definition 3.1.8.
Ad (5). Since P is an orthocomplemented poset,
there exists OEP such that 0 9 b for all bEP. Hence, for
any Psequence y, 0 b' for all bEB Therefore
fb(0) = g (0) = 0 for all [y]EE and for all bEB .
Ad (6). For any eEP such that e X 0, there
exists a Fsequence y such that g (e) = 1 by Condition II.
Hence sup g (e) = 1. Since this holds for all nonzero
eEP, it follows that inf (sup (g (e))) = 1 > .
eEP [y]EZ Y
e#O
3.3.6 Corollary. Let P satisfy Condition II, let <* be
the "weak" partial order induced on P by E (cf. Definition
3.3.1), and let e,fEP. Then e : f if and only if e d* f.
Proof. By Lemma 3.1.14 e s f if and only if
87
g (e) : g (f) for all [y]EZ. But by Definition 3.3.1,
since the T of Definition 3.3.3 is a Travis matrix for
(2,P,'), g (e) g (f) for all [y]EZ if and only if e :* f.
Therefore e f if and only if e !* f.
4. Examples
We now give deterministic Travis matrices which
correspond to some of the posets previously mentioned.
Figures 12, 13, 14, 15, and 16 are deterministic Travis
matrices for the posets given in Figures 1. 3. 5, 8, and
10, respectively. Figure 17 is the Hasse diagram of the
orthomodular post which is not a lattice, given by
Janowitz [5] and mentioned in Example 2.1.24; Figure 18 is
a deterministic Travis matrix for it.
0 e f 1 e' f'
S 0 1 1 1 0 0
S2 0 0 1 1 1 0
S3 0 0 0 1 1 1
Figure 12
Figure 13
a' b' c' d' e'
0 1 1 0 1
0 1 1 1 0
0 1 1 1 1
1 0 1 0 1
1 0 1 1 0
1 0 1 1 1
1 1 0 1 1
1 1 1 0 1
1 1 1 1 0
1 1 1 1 1
Figure 14
0
S1 0
S2 0
S3 0
S4 0
S5 0
S6 '0
S7 0
Ss
Sg 0
S9 0
S10 0
a b
1 0
1 0
1 0
0 1
0 1
0 1
0 0
0 0
0 0
0 0
c d e
0 1 0
0 0 1
0 0 0
0 1 0
0 0 1
0 0 0
1 0 0
0 1 0
0 0 1
0 0 0
I  
_
0 a b c e f q 1 a' b' c' e' f' q'
S 0 1 0 0 1 0 0 1 0 1 1 0 1 1
S2 0 1 0 0 0 1 0 1 0 1 1 1 0 1
S3 0 1 0 0 0 0 1 1 0 1 1 1 1 0
S4 0 1 0 0 0 0 0 1 0 1 1 1 1 1
S5 0 0 1 0 1 0 0 1 1 0 1 0 1 1
S6 0 0 1 0 0 1 0 1 1 0 1 1 0 1
S7 0 0 1 0 0 0 1 1 1 0 1 1 1 0
S8 0 0 1 0 0 0 0 1 1 0 1 1 1 1
S9 0 0 0 1 1 0 0 1 1 1 0 0 1 1
S10 0 0 0 1 0 1 0 1 1 1 0 1 0 1
S11 0 0 0 1 0 0 1 1 1 1 0 1 1 0
S12 0 0 0 1 0 0 0 1 1 1 0 1 1 1
S13 0 0 0 0 1 0 0 1 1 1 1 0 1 1
S14 0 0 0 0 0 1 0 1 1 1 1 1 0 1
S15 0 0 0 0 0 0 1 1 1 1 1 1 1 0
S16 0 0 0 0 0 0 0 1 1 1 1 1 1 1
Figure 15
Figure 16
0 a b c d e f g 1 a' b' c' d' e' f' q'
S1 0 1 0 0 1 0 1 0 1 0 1 1 0 1 0 1
S2 0 1 0 0 1 0 0 1 1 0 1 1 0 1 1 0
S3 0 1 0 0 1 0 0 0 1 0 1 1 0 1 1 1
S4 0 1 0 0 0 1 0 0 1 0 1 1 1 0 1 1
S5 0 1 0 0 0 0 1 0 1 0 1 1 1 1 0 1
S6 0 1 0 0 0 0 0 1 1 0 1 1 1 1 1 0
S7 0 1 0 0 0 0 0 0 1 0 1 1 1 1 1 1
S8 0 0 1 0 1 0 1 0 1 1 0 1 0 1 0 1
S9 0 0 1 0 1 0 0 1 1 1 0 1 0 1 1 0
S1 0 0 1 0 1 0 1 0 0 1 1 0 1 0 1 1 1
S11 0 0 1 0 0 1 0 0 1 1 0 1 1 0 1 1
S12 0 0 1 0 0 0 1 0 1 1 0 1 1 1 0 1
S13 0 0 1 0 0 0 0 1 1 1 0 1 1 1 1 0
S14 0 0 1 0 0 0 0 0 1 1 0 1 1 1 1 1
S15 0 0 0 1 0 0 1 0 1 1 1 0 1 1 0 1
S6 0 0 0 1 0 0 0 1 1 1 1 1 1
S17 0 0 0 1 0 0 0 0 1 1 1 0 1 1 1 1
S18 0 0 0 0 1 0 1 0 1 1 1 1 0 1 0 1
S19 0 0 0 0 1 0 0 1 1 1 1 1 0 1 1 0
S20 0 0 0 0 1 0 0 0 1 1 1 1 0 1 1 1
S21 0 0 0 0 0 1 0 0 1 1 1 1 1 0 1 1
S22 0 0 0 0 0 0 1 0 1 1 1 1 1 1 0 1
S23 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 0
S24 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1
Figure 17
Figure 18
0 a b c d e f q h a' b' c' d' e' f' q' h'
S1 0 1 1 0 0 0 0 0 0 1 0 0 1 1 1 1 1 1
S2 0 1 0 0 1 0 1 0 0 1 0 1 1 0 1 0 1 1
S3 0 1 0 0 1 0 0 0 0 1 0 1 1 0 1 1 1 1
S4 0 1 0 0 0 0 1 0 0 1 0 1 1 1 1 0 1 1
S5 0 1 0 0 0 0 0 0 0 1 0 1 1 1 1 1 1 1
S6 0 0 1 1 0 1 0 0 0 1 1 0 0 1 0 1 1 1
S7 0 0 1 1 0 0 0 0 0 1 1 0 0 1 1 1 1
Sg 0 0 1 0 0 1 0 0 0 1 1 0 1 1 0 1 1 1
S9 0 0 1 0 0 0 0 0 0 1 1 0 1 1 1 1 1 1
S10 0 0 0 1 1 1 1 0 0 1 1 1 0 0 0 0 1 1
S11 0 0 0 1 1 1 0 0 0 1 1 1 0 0 0 1 1 1
S12 0 0 0 1 1 0 0 0 0 1 1 1 0 0 1 1 1 1
13 0 0 0 1 1 0 1 0 0 1 1 1 0 0 1 0 1 1
S14 0 0 0 1 1 0 0 1 0 1 1 1 0 0 1 1 0 1
S15 0 0 0 1 0 1 0 0 0 1 1 1 0 1 0 1 1 1
S16 0 0 0 1 0 1 1 0 0 1 1 1 0 1 0 0 1 1
S17 0 0 0 1 0 0 1 0 0 1 1 1 0 1 1 0 1 1
S18 0 0 0 1 0 0 0 1 0 1 1 1 0 1 1 1 0 1
S19 0 0 0 1 0 0 0 0 0 1 1 1 0 1 1 11 1
S20 0 0 0 0 1 1 1 0 0 1 1 1 1 0 0 0 1 1
S21 0 0 0 0 1 1 0 0 0 1 1 1 1 0 0 1 1 1
s22 0 0 0 0 1 0 1 0 0 1 1 1 1 0 1 0 1 1
S23 0 0 0 0 1 0 0 1 0 1 1 1 1 0 1 1 0 1
524 0 0 0 0 1 0 0 0 0 1 1 1 1 0 1 1 1 1
s25 0 0 0 0 0 1 1 0 1 1 1 1 1 1 0 0 1 0
S26 0 0 0 0 0 1 1 0 0 1 1 1 1 1 0 0 1 1
27 0 0 0 0 0 1 0 0 1 1 1 1 1 1 0 1 1 0
S28 0 0 0 0 0 1 0 0 0 1 1 1 1 1 0 1 1 1
S29 0 0 0 0 0 0 1 0 1 1 1 1 1 1 1 0 1 0
S30 0 0 0 0 0 0 1 0 0 1 1 1 1 1 1 0 1 1
31 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 0 0
S32 0 0 0 0 0 0 0 1 0 1 1 1 1 1 1 1 0 1
S33 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 0
S34 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1
