CD I')I)() Birkh 川 IScr Vcrlag. Ba~cl. In Memory o[ Evelyn Ne/son. OML L has the Block Exlensioll. Property. Algebra Universalis, 27 (1990) 1-9

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1 Algebra Universalis, 27 (1990) W/'XI/O 10m 1-1 附 $ υ/0 CD I')I)() Birkh 川 IScr Vcrlag. Ba~cl 810cks and commutators in orthomodular lattices GÜNTER BRUl 吨 S AND RICHARD GREECHlE In Memory o[ Evelyn Ne/son An orthomodular lattice (abbreviated: OML) is block-finite, [2], if thc sct ~(/_ of all its blocks is finite, it is comml 归 tor-finite, [ 坷, if the set com L of all its commutators is 白 nite. The class of all commutator-finite OMLs properly contains the class of all block-finite OMLs, see [5]; and, every finitely generatcd block-finite OML is finite, see [1]. The question thus arises whcthcr evcry finitely generated commutator- 自由 te OML is also finite. Wc answcr this qucstion affirmatively in this paper. For this purpose we introduce a new conccpt, thc BIock Extension Property. We say that an OML L has the Block Extensìon Property if for cvcry finitc sct m of blocks of L there exists a block-finite full subalgebra S of L containing U 坦. A subalgebra S of L is said to be [u /l if every block of S is a block of L. Thc main result of this paper is: THEOREM. Every commutator 尹 nîte Property. OML L has the Block Exlensioll We, in fact, construct an extension S in such a way that com S == com L. Sincc every finite subset of a commutator finite OML is trivially containcd in thc union of finitely many blocks, it follows from our Theorem that it is containcd in a block-finite subalgebra. Hence, by [1], the Sl 归 1gebra gcncratcd by it is finitc. We thus obtain: COROLLARY. Every finìtely generated commutator-finite OML ìs finite. In the first section we generalize two results concerning paths in OMLs, which were proved in [2] for block-finite OMLs, to fit our present needs. In the second part we describe how blocks can be constructed from horizontal summands of intervals. We also state a general form of an Excìsion Lemma. A specification of it Presented by Alan Day. Received January 7, 1988 and in lìnal form July 5, 1988.

2 2 GûNTER BRUNS AND RICHARD GREECHIE ALGEBRA UNIV. is given in the next sections. It is the main tool used to prove the above theorem in section 4. In the concluding section we mention some open problems connected with our investigations. We are grateful to L. Herman for stimulating conversations concerning the subject matter of this paper. We also gratefully acknowledge direct and indirect support by the Natural Science and Engineering Research Council of Canada. 1. Paths In this section we show that two theorems concerning paths in OMLs, (4.4) and (4.5) of [2], which were proved in [2] for block-finite OMLs, hold under a weaker assumption, the second in a slightly weakened form. Both generalizations will be needed 仕 equently in this paper. We first recall some relevant definitions and facts, see [2, 3, 4, 6]. The commutator x * y of elements x, y of an OML is defined by x * y = (x v y) ^ (x v y') ^ (x' v y) ^ (x' v y'). If the union of two distinct blocks (maximal Boolean subalgebras) A, B of L is a subalgebra of L then there exists a unique commutator e E A n B such that A n B = (A U B) n Se, where Se = [0, e'] U [e, 1]. We express this fact briefly by A - e B or, if e is irrelevant, by A - B. A path is a sequence Bo, B 1,, Bn (n 三 0) of blocks such that Bi - B 川 holds whenever 0 三 i < n. The path is said to join the blocks Bo and B n- The path is said to be strictly proper if Bi n B i+ l 手 C(L) holds for all i, it is said to be proper if it is strictly proper or if n = 1. The OML L is defined to be path-connected if any two blocks of L can be joined by a proper path. THEOREM 1. Every commutator-finite OML L is path-connected. Proof by induction on n = Icom LI. If n = 1, L is Boolean and the claim is trivially true. Assume n ~ 2. By Theorem 14 of [5] and the fact that the product of two path-connected OMLs is again path-connected, (4.3) of [ 巧, we may assume that L is irreducible. Let A, B be distinct blocks of L. Assume first that AnB 学 {O, 1}. We show that in this case A and B can be joined by a strictly proper path. Choose a ε AnB, a =Þ O, 1. If comc(a) 手 com L then, by inductive hypothesis, A and B can be joined by a proper path in C(a), and hence, since a ft C(L), by a strictly proper path in L. We may thus assume that com C(a) = com L, in particular that V com C(a) = V com L = 1. This implies by Theorem 4

3 Vol. 27, 1990 Blocks and commutators in orthomodular lattices 3 of [5] that C(a) has no non-trivial Boolean factor. But C(a) = [0, a] X [0, a ']. Since [0, 叶, [0, a'] are not Boolean they have fewer commutators than C(a), hence are path-connected by inductive hypothesis. Thus A, B can be joined by a strictly proper path in L. The case A n B = {O, 1} reduces to the previous one as in (4.4) of [2]. An OML L is the horizontal sum of a family (L;);EI of at least two subalgebras, see [2, 7], if U L; = L, L; 手 Lj and L; n Lj = {O, 1} whenever i 手 j, and one of the following equivalent conditions is satisfied. The three conditions relevant for our purposes are: 1. If x ε L; 一 Lj, Y E Lj 一 L; then x v y = 1; 2. Every block of L belongs to some L;; 3. If S; is a subalgebra of L; then υ S; is a subalgebra of L. THEOREM 2. Let L be a path-connected OML without non-trivial Boo/ean factor which contains at least two blocks which cannot be joined by a strictly proper pa 仇 Dφne a relation 三 in 现 by A == B ~ A and B can be joined by a strictly proper path. Then == is an equivalence relation in 现 L. 扩 íb is an equivalence class of 21 L modulo 三 then U íb is a subalgebra of L with 21u 恐 = íb and L is the horizontal sum of these subalgebras. Proof. If L were the direct product of non-degenerate OMLs L 1 and L 2 then, considering how the paths in L are related to the paths in the factors, see [2], it would follow that any two blocks in L 1 and any two blocks in L 2 could be joined bya (not necessarily proper) path. Thus any two blocks in L could be joined by a strictly proper path, contrary to the assumptions of Theorem 2. Thus L is irreducible. To complete the proof first note the following: ifa, B ε 现 L anda nb 学 {O, 1} then A 三 B. (*) For, A and B may be joined by a proper path. If one of these paths has length n 三 2 it is strictly proper and hence A 三 B. If n = 1 then the path A ~ B is strictly proper since A n B 手 {O, 1} = C(L) and again A 三 B, proving (*). Now let íb be an equivalence class of 现 L modulo 三 and let a, b E U íb. We show α vb εu íb. This is clear if a ε{o, 1}. We may thus assume that a 学 0, 1. There exist blocks Aε íb and B ε 也 L such that a ε A and a, a v b ε B. Since a E A n B it follows from (*) that B E 58 and hence a v b εu 埠, making U íb a subalgebra. The rest of the claim follows easily from (*).

4 4 GUNTER BRUNS AND RICHARD GREECHIE ALGEBRA UNIV We do not know whether Theorem 2 remains true if the assumption that L has no Boolean factor is dropped and "horizontal sum" in the conclusion is replaced by "weak horizontal sum," see [2]. This would be true if eve 可 path-connected OML had a largest Boolean factor. We do not know whether this is the case. We will frequently use the following consequence of Theorem 2. COROLLARY 1. 扩 L is a non-boolean path-connected OML then eveη block A o[ L which is not a horizontal summand o[ L contains a commutator U 手 0, 1. Proof Since L is not Boolean and path-connected there exists a proper path A-vB- in L. If v 学 1 the claim is proved. If v = 1 then, by Theorem 4 of [ 匀, L has no Boolean factor. If A could not be joined with any other block by a strictly proper path it would be a horizontal summand by Theorem 2. Thus there exists a strictly proper path A - v B... and v 手 0, Blocks and subalgebras For the rest o[ the paper we assume that L is an OML every subalgebra of which is path-connected. Only Lemma 2 and the Excision Lemma do not require this assumption. If A ç; L, u ε L we define u ^ A = {u ^ X I X ε A }. It is well-known that if A is a block of L and u ε A then u ^ A = [0, u] na is a block of the OML [0, u]. LEMMA 1. 扩 U εaε 现 L and A n com L ç; {O, u}, then [0, u] is Boolean or u ^ A is a horizontal summand o[ [0, u]. Proof Since u ^ A contains no commutators 手 0, u the claim follows from Corollary 1. COROLLARY 2. 扩 u is minimal in (A n com L) 一 {O} then u 八 A is a horizontal summand o[ [0, u]. Proo[. Since u εcom [0, 叶, Lemma 1. [0, u] is not Boolean and the claim follows from For X 1, X 2,..., X n ç; L define?1xi=(ylxz Xt 均 )

5 Vol. 27, 1990 Blocks and commutators in orthomodular lattices 5 LEMMA 2. Let { 屿, 屿,..., Un} be an orthogonal subset of L with V7=t U; = 1 and for eveη i let A; be a block of [0, U;]. Then V7=t A; is a block of L. We leave the easy proof of this to the reader. LEMMA 3. Let L be commutator-finite, A ε~ll and let { 屿, 屿,..., un} be a maximal orthogonal subset of the set of all minimal elements of (A n com L) 一 {O}. D e, 卢 ne Un+l = 八乙 1 U;. Then, for 1 三 i 三 n, U; ^ A is a Boolean horizontal summand of [0, u;], [0, un+d is Boolean and A = V7:/ (u; ^ A). Proof. The first claim restates Corollary 2. Since { 屿,..., un} is maximal, Un+l ^ A contains no commutator v 手 O. In particular, Un+l ft com L and hence [0, un+d is not a horizontal sum. Thus [0, un+d is Boolean by Lemma 1. By Lemma 2, V?:l (u; ^ A) is a block which trivially contains A, proving the last claim. A key step in the proof of our main theorem is an application of a specification of the fo l1owing general result. We write M 三 L to indicate that M is a subalgebra of L. EXCISION LEMMA. Let M 三 L and WçW L. 扩 U(W L -W)nUWçM then (L-U 现 )UM 三 L. Proof. Let L 1 =(L-UW)UM and x, y ε L 1 Since L 1 is closed under orthocomplementation we need only show that x v y ε L 1 Suppose not; then xvy ε(u W) - M, so, by hypothesis, x v y 1$ U ( 也 L 一刻 ). Hence every block containing x v y belongs to 现. Since x, x v y belong to some block, x εu ~l. By symmetry, yεu 现. Thus, by hypothesis, x, y E M and hence x v y ε M, completing the proof. 3. Inessential blocks of intervals For u ε L let ß(u) be the set of all Boolean horizontal summands of [0, u]. Note that 现 [O.u] 一 {X} 学臼 whenever X ε 卢 (u). An Xε 卢 (u) is said to be essential if there exists D ε 现 L such that u 1$ D and X n D 手 {O}, and inessential otherwise. If X E W[o.u] define Lx = (L - C(X)) U Su. We shall show that if u E com L - {O} and X ε 卢 (u) then Lx is a subalgebra of L iff X is inessential. The subalgebras Lx play a crucial role in the sequel. We leave the proof of the following simple lemma to the reader.

6 6 GÜNTER BRUNS AND RICHARD GREECHIE ALGEBRA UNJV LEMMA4. 扩 uε L and if X is a block of [0, u] then (1) C(X) = C(u) 门 {x ε L lu ^x ε X}, (2) Lx = (L - C(u)) U Su U {x ε L I u ^x ~X}, (3) X n Lx = {O, u} and L 一 Lx = C(X) - Su c:; C(u) 一乱, (4) [O, u]-x c:; Lx. LEMMA 5. Let u ε coml 一 {O} and X ε 卢 (u). X is inesseωial i 万 Lx 三 L Proof Assume that X is inessential. We apply the Excision Lemma with 现 = ~c(x) and M = Su. Since Su 三 L we need only verify that (U ( 现 L 一 2l c (x))) C(X) c:; Su. To this end, let x ε A n B where A ε 2lc(X) and B E 现 L 一 2l C( x). Assume u ~ B. Then there exist y ε B such that y (þ u. Since x v y, x' v y Cu would imply y C u we may assume that x v y (þ u. Then there exists a block D containing u ^ x and x v y and hence not containing u. Since X is inessential and u ^x ε X, it follows that u ^ x = 0; moreover, since x C u, we obtain x ε Su. W 巳 may thus assume that u ε B. There exists a path A = Ao ~ AJ ~... ~ An = B in C( 杠, u}). Let j be the smallest index such that u ^ Aj+l 学 X. Then u ^ A j = X and (u^aj+ 1 )nx={0, u}. Let Aj~vAj+l. Then v$u or u 三 v'. But u 三 v' would imply X c:; Aj+l' contrary to our assumption. Thus v 三 u, hence u = v. It follows that x ε Aj n Aj+l c:; Su- To prove the converse assume D ε 现 L and O 手 xεd n x. We have to show that u εd. Suppose that u ~ D so that there exists y εd, y (þ u. Since y = (y v x) 八 (y v x') and u' 三 y v x' C u it follows that y v x <t u. Since x 三 u ^ (x v y) < u and X ε 卢 (u) we obtain u 八 (x v y) ε X. But U E Lx and, since x v y <t u, x v y ε Lx by Lemma 4. Thus, by hypothesis, U^(xvy) ε Lx, contradicting X 门 Lx = {O, u}. lt follows that u ε D, completing the proof. COROLLARY 3. 扩 Xε 卢 (u) is inessential then Lx is the largest subalgebra N of L such that X n N = {O, u}. Proof. By Lemmas 4 and 5, Lx has the property. Let N ::, 三 L with xnn== {O, u} and suppose n εn 一 Lx. By Lemma 4, n ε C(X) 一乱, so that 0 < u ^ n < u. But u ^ n ε N n C(X) n [0, u] = N n X = {O, u}, a contradiction. Hence N ç Lx and Lx is the largest such subalgebra. LEMMA6. 扩 0 手 U ε coml and 扩 Xε 卢 (u) is inessential then (1) if y <t u then Sy c:; Lx, (2) com L c:; Lx, (3) 扩 u 手 uε coml 一 {O} and Y ε ß(v) then Y c:; Lx, (4) 扩 uεcoml and 扩 [0, 叫 is Boolean then [0, v'] c:; Lx. n

7 Vol. 27, 1990 BIocks and commutators in orthomodular latticcs 7 Proof. Assume that there exists z ε L such that u *y 三 z f$ Lx. Then 0 < u ^ z' 三 u ^y' <u. Since u ^ z' ε X it would follow that u ^ y' ε xn Lx, contradicti 吨 ) of Lemma 4. Thus [y, 1] ç Lx and hence Sy ç Lx, proving (1). To prove (2) assume v εcom L and v f$ Lx. Then v ε C(u) 一乱, O<U^V, U^V'<u and u ^ V, U ^ V' ε X. Since [0, u ^ v] is Boolean, u 八 v is not a commutator and it follows from Corollary 6 of [5] that C(v) cþc(u). Thus there exists B ε ~(L such that u f$ B and v ε B. Let u, v εaε W L and let A =Ao-A 1 - -An=B be a path in C( v). Let p be an index such that u ε Ap - AP+l and Ap -w AP+l' Since v ε Sw, u ^ v or u ^ v' is in X nap+l' contradicting the fact that X is inessential and proving (2). To prove (3) assume that under the assumptions of (3) there exists y εy 一 Lx. Then, by (2) of Lemma 4, 0 < u ^ Y < u and u ^y ε X. Since u ^ y 三 u ^ V :5 U this implies u ^ v ε X which, since v ε Lx, gives u < v. But 0 < u ^ Y 三 y < v and y ε Y implies u ^ y ε Y. This with u ^ y 三 u < v implies u ε Y and the Boolean algebra [0, u] contains a commutator, a contradiction proving (3). Assume fir 时 ly that v satisfies the assumptions of (4) and there exists y 三 v', Y I$ Lx. Then o<u ^y 运 u ^ v' 三 u and u ^y ε X. Thus also u ^ v' ε X and, since v ε Lx by (2), u 三 v', contradicting the fact that [0, υ'] is Boolean. 4. Proof of the main theorem Assume 0 学 U ε coml and X ε 卢 (u). We say that a commutator v is associated with X iff There exist A, E ε W L such that u EA - E, u 手 v, A -tj E andx ça. (*) LEMMA 7. 扩 O 手 U εcom L and if X ε 卢 (u) is essential then there exists α commutator v associated with X. 扩 X :;é Y ε 卢 (u) and Y is also essential then the sets of commutators associated with X and Y are disjoint. Proof. Since X is essential there exists D ε ~ll such that u 1$ D and X n D 学 {O}. Let Ao E 现 L contain X and let Ao - A An = D be a path in C(X n D). There exists p such that u ε Ap - Ap+l' Ap -u Ap+l' Put A = A l" E=Ap+l' There exists x>o, x ε xnd ça ne. But v 三 x would imply v 三 u, hence u ε Ap+l' a contradiction. Thus x 三 u^ υ, and u 非 v. Since x ε xna, x < u (since u 1$ D) and u ^ A is a block in [0, u] it follows that u ^ A = X ç A. Thus v is associated with X. To prove the second part let v be associated with X and w with Y. Then there exist blocks A, B as above such that u, v ε A, u, w ε B, X ça and Y ç B. It follows that u ^ v' ε X and u ^ w' ε Y. It follows that u ^ v' 学 u ^ w' and hence v 手 w, proving the second part.

8 8 GUNTER BRUNS AND RICHARD GREECHIE ALGEBRA UNIV As an immediate consequence of this we obtain the following. COROLLARY 4. Commutator-finite OMLs contain only finitely many essential Boo/ean horizontal summands of intervals [0, u] with u ε com L. COROLLARY 5. A commutator 乔 nite OML is not b/ock 抖 nite 扩 and only 扩 it contains an interval having infinitely many inessential Boo/ean horizonta/ summands. The last corollary, together with the applicaion of the Excision Lemma which follows, exhibits a structural feature of commutator-finite OMLs which shows how closely related they are with block-finite OMLs and, by the corollary to the main theorem, with finite OMLs. We are now in a position to prove the main Theorem. Let L be commutatorfinite and 58 a finite set of blocks of L. For every u εcom L choose a u, b u such that u = a u * b u and add a block to 现 containing a u and a block containing b uo Then 现 remains finite and com S = com L holds for every subalgebra S of L containing U 58. For every u ε coml, u 学 0, let Qu be the set of all those Boolean horizontal summands of [0, u] which are either essential or of the form U 八 A for some A ε58 with u ε A. By Corollary 4 and the 自 niteness of ØJ each Qu is finite. Define Q~ = ß(u) - Qu. Define s=n{lxio 手 uε coml, X ε Q~} Clearly S is a subalgebra of L. By (4) of Lemma 4 and (3) of Lemma 6 we have u Qu ~ S whenever 0 手 U E com L. By Lemma 3 and (4) of Lemma 6 it follows that U 58 ~ S. We prove next that every block A of S is a block of L, i.e., that the subalgebra S of L is full. Extend A to a block A of L. Since com S = com L the commutators in A and A are the same. Let u be a minimal element of (A n com L) - {O} = (A n com L) 一 {O}. By Corollary 2, u ^ A is a Boolean horizontal summand of [0, u] n S and u ^ A is a Boolean horizontal summand of [0, u] n S and u 八 A is a Boolean horizontal summand of [0, u]. In particular there exists x ε u ^ A, 0 < x < u. It follows from (3) of Lemma 4 that u ^ A ε Qu and hence u ^ A ç S. Since u 八 A is a block in [0, u] n S it follows that u ^ A = u ^ A and therefore u ^ A ε Qu. Now let { 屿, 屿,..., u n } be a maximal orthogonal subset of the set of all minimal elements of (A n com L) 一 {O} and let v=v 二 1 Ui. By Lemma 3 [0, 叫 is Boolean; by Corollary 6 of [5] and (4) of Lemma 6 it is contained in S so that [0, v'] ç A n S = A. It now follows from Lemma 2 and 3 that A = A and hence that S is a full subalgebra. Since all Qu are finite it also follows that S is block-finite, completing the proof.

9 Vol. 27, 1990 Blocks and commutators in orthomodular latticcs 9 5. Concluding remarks It is easy to construct examples of OMLs which have the Block Extension Property (abbreviated: BEP) but are not commutator-finite. It would be of interest to find classes of OMLS having the BEP which properly contain the class of commutator-finite OMLs or, ideally, to find alternative descriptions of OMLs having the BEP. The BEP is trivally preserved under finite products. Which other basic algebraic constructions preserve the BEP? In particular, is the BEP preserved by subalgebras and homomorphic images? The BEP has as an obvious consequence that every subalgebra generated by the union of finitely many blocks Ìs block-finite. We call this last statement the Weak Block Extension Property, abbreviated: WBEP. The WBEP is obviously enough to ensure that every finitely generated subalgebra is finite. How is the WBEP related to the BEP? Is it strictly weaker or equivalent? It is easy to see that the WBEP is preserved under finite products and subalgebras. Is it preserved under homomorphic images? We believe that answers to such questions would prove signi 且 cant in the developing theory of orthomodular lattices. REFERENCES [1] G. BRUNS, A finiteness criterion for orthomodular lattices, Can. J. Math. 30 (1978), [2] G. BRUNS, Block-finite orthomodular lattices, Can. J. Math. 31 (1979) [3] D. J. FOUL 邸, A note on orthomodular lattices, Portugal Math 21 (1966), [4] R. J. GREECHIE, On generating pathological orthomodular structures, Kansas State Univ. Technical Report 13 (1970). [5] R. GREECHIE and L. HERMAN, Commutator-finite orthomodular lattices, Order 1 (1985), [6] G. KALMBACH, Orthomodular lattices, Academic Press (1983). [7] M. D. MACLAREN, Nearly modular orthocomplemented lattices, Boeing Sci. Res. Lab D (1964). Department of Mathematics McMaster University Hamilton, Ontario L8S 4Kl Kansas State University Manhattan, Kansas U.S.A., 66506