(Positive) Totally Ordered Noncommutative Monoids -- How Noncommutative Can They Be?
aa r X i v : . [ m a t h . L O ] M a y (POSITIVE) TOTALLY ORDERED NONCOMMUTATIVE MONOIDS –HOW NONCOMMUTATIVE CAN THEY BE? ELIAHU LEVY
Abstract.
Commutative totally ordered monoids abound, number systems for example.When the monoid is not assumed commutative, one may be hard pressed to find an example.One suggested by Professor Orr Shalit are the countable ordinals with addition.In this note we attempt an introductory investigation of totally (also partially) orderedmonoids, not assumed commutative (still writing them additively), and taking them as pos-itive, i.e. every element is greater than the unit element. That, in the usual commutativecases, allows the ordering to be defined via the algebraic structure, namely, as divisibility (inour additive sense): a ≤ b defined as ∃ c ( b = a + c ). The noncommutative case offers severalways to generalize that.First we try to follow the divisibility definition (on the right or on the left). Then, alterna-tively, we insist on the ordering being compatible with the operation both on the left and onthe right, but strict inequality may not carry over – again refer to the ordinals example. Wetry to see what axiom(s) such requirements impose on the monoid structure, and some factsare established.Focusing especially on the totally ordered case, one finds that necessarily the noncommu-tativity is somewhat limited. One may partly emulate here the commutative case, speakingabout infinitely grater vs. Archimedean to each other elements, and in the Archimedean caseeven emulate Euclid’s Elements’ theory of ‘ratios’ – all that imposing some partial commuta-tivity. Introduction
One encounters many examples of totally ordered commutative monoids M (which we shallwrite additively). They have a unit element (denoted by 0), and have the set M + of thenonnegative elements, which is a submonoid, and the ordering may be defined via M + : b ≥ a ≡ ∃ c ∈ M + ( b = a + c ). Further restricting ourselves to M + , i.e., assuming M = M + , werefer to the monoid as positive . Then(1) b ≥ a ⇔ ∃ c ( b = a + c ) . I.e., speaking multiplicatively, b ≥ a means b is divisible by a .And this is assumed to be a total ordering: transitive (as (1) implies, using the associativityof the monoid) but also antisymmetric and total: for a = b , one and only one of a > b or b > a holds. Of the many examples one may cite the positive integers (or rationals, reals)But what if the monoids is noncommutative? then even in (1) there may be a differencebetween requiring divisibility in the left or on the right (or maybe both?). Also, one wouldbe hard pressed to find an indeed noncommutative example.An example, as noted by Professor Orr Shalit, are the ordinals (say the countable ordinals,to make them a set and not a proper class), with respect to addition . Here indeed addition isassociative, but not commutative: ω + 1 = ω = 1 + ω , and the ordering is characterized by Date : May 2020.
Key words and phrases. monoid, noncommutative, totally ordered, partially ordered, free noncommutativemonoid, idempotents, Archimedean elements, Infinitely small, Euclid Elements’ ‘ratios’. (1) in that order: b = a + c . Indeed, by the theory of ordinals (and well-ordered sets), b ≥ a means that a is an initial segment of b , and c will be what b has after that initial segment.Note that here we have compatibility of the ordering: a ≤ b ⇒ c ′ + a ≤ c ′ + b , which ofcourse follows from the characterization (1) (and associativity), but also compatibility on theright: a ≤ b ⇒ a + c ≤ b + c , but here strict inequality would not be carried over: 0 < ω = ω = 1 + ω .In this note it is attempted to shed some introductory light on these issues (with the orderingpartial or total).In the first section we try to follow the divisibility ‘paradigm’ in the noncommutative case,assuming cancelation (both on the left and on the right), compare [3] Ch. 4. We show byexample that one may still have then both noncommutativity and a divisibility ordering whichis total.Next, alternatively, one insists on the ordering being compatible with the operation bothon the left and on the right, embarking on a somewhat methodical investigation about whataxiom(s) would be imposed by our requirements on the monoid structure, proving some facts.We focus then on the possibilities for idempotents and an element a absorbing an element b on the left or on the right (i.e. a + b = a or b + a = a ) in positive partially ordered monoids –one does not find them when there is cancelation, but in the general case they may play somepart.Finally the totally ordered case is considered. There one finds that necessarily the noncom-mutativity is somewhat limited. One may partly emulate here the commutative case, speakingabout infinitely greater vs. Archimedean to each other elements, and in the Archimedean caseeven emulate Euclid’s Elements’ theory of ‘ratios’ – all that imposing some partial commuta-tivity.1.1. Some Notation. M is a monoid (written additively, though not assumed commutative),i.e., a set with an associative operation + having a unit element 0. We endow it with anordering relation ≤ (whose inverse relation we, as usual, denote by ≥ ). We consider thefollowing postulates that may (or may not) hold:(O) Transitive: If a ≤ b and b ≤ c then a ≤ c, ∀ a, b, c ∈ M. (P) Positive: a ≥ , ∀ a ∈ M. (C) Compatible: If a ≤ b then a + c ≤ b + c and c + a ≤ c + b, ∀ a, b, c ∈ M. (T) Total: For any a, b ∈ M, either a ≤ b or b ≤ a (allowing both) . (A) Antisymmetric: For a, b ∈ M, if both a ≤ b and b ≤ a then a = b. Definition 1.
A monoid endowed with an ordering relation satisfying (O), (P), (C) as aboveis called a positive preordered monoid . If (A) is also satisfied, a positive partiallyordered monoid , and if both (A) and (T) are also satisfied, a positive totally orderedmonoid . Ordering Defined by ‘Divisibility’ in Monoids With Cancelation
Let M be a (possibly) noncommutative monoid. We assume cancelation on the right andon the left a + c = b + c ⇒ a = b, c + a = c + b ⇒ a = b. That would be, of course, a prerequisite for any possibility to embed M into a group. POSITIVE) TOTALLY ORDERED NONCOMMUTATIVE MONOIDS 3
Definition 2.
The right divisibility preorder (resp. left divisibility preorder ) on M isdefined as (2) a ≤ r b : ⇔ ∃ c ( b = a + c ) (resp. (3) a ≤ l b : ⇔ ∃ c ( b = c + a )One clearly sees that the associativity of the operation implies these are indeed (positive)preorders – satisfy (O) and (P), and each satisfies one of the assertions of (C): for rightdivisibility a ≤ r b ⇒ c ′ + a ≤ r c ′ + b , while for left divisibility a ≤ l b ⇒ a + c ′ ≤ l b + c ′ . Proposition 3.
The following are equivalent: (1) M with right divisibility is a partial ordering, i.e. satisfies (A). (2) M with left divisibility is a partial ordering, i.e. satisfies (A). (3) If a sum a + . . . + a n equals then all the summands are . Proof . (1) implies (2) (and similarly (1) implies (3)). Indeed, in (3)0 = a + . . . + a n ≥ r a + . . . + a n − ≥ r · · · ≥ r a ≥ r
00 = a + . . . + a n ≥ l a + . . . + a n ≥ l · · · ≥ l a n ≥ l . On the other hand, assume (3) holds.Now, a ≤ r b ≤ r a means ∃ c, c ′ such that b = a + c , a = b + c ′ . Then a + 0 = a = a + c + c ′ and by cancelation c + c ′ = 0, hence by (3) c = c ′ = 0, thus a = b . Similarly for ≤ l . (cid:3) Extending the Monoid by ‘Annexing’ Differences (Assuming the Ordering ≤ r Total).
By cancelation the c in Def. (2), when it exists, is unique.Thus if a ≤ r b , we write the unique c satisfying b = a + c as − a + b .And if a ≤ l b , we write the unique c satisfying b = c + a as b − a .(Note that − ( ) is used here as ( ) − is used for multiplicative notation – minding the orderof summands. In particular − ( a + b ) = ( − b ) + ( − a )!)And in analogy with the way one formally extends the natural numbers to the integersor rationals, we may try to ‘annex’ differences b − a . Yet with noncommutativity complica-tions may well arise. (Compare [2] Ch. 7 about quotient noncommutative rings and the Orecondition.)To fix matters, suppose the ordering ≤ r is total .Then define the extension by differences of M as the set M × M , a pair ( a, b ) therewritten as b − a , modulo the equivalence relation (stemming from an identity ( b + c ) − ( a + c ) = b − a valid in some embedding group, say):(4) b − a ≡ b − a ⇔ ∃ c, c ′ ( b + c = b + c ′ , a + c = a + c ′ ) . That factor set (where we denote by [ b − a ] the equivalence class of b − a ) is endowed by theoperation defined by:(5) [ c − b ] + [ b − a ] = [ c − a ] . (5) makes sense. Indeed, to find [ b − a ] + [ b − a ],one either has a ≤ r b , then there is c = − a + b , a + c = b and[ b − a ] + [ b − a ] = [( b + c ) − ( a + c )] + [ b − a ] = [( b + c ) − a ] :(6) [ b − a ] + [ b − a ] = [( b + ( − a + b )) − a ] if a ≤ r b ELIAHU LEVY or a ≥ r b , then there is c ′ = − b + a , a = b + c and[ b − a ] + [ b − a ] = [ b − a ] + [( b + c ) − ( a + c )] = [ b − ( a + c )] :(7) [ b − a ] + [ b − a ] = [ b − ( a + ( − b + a ))] if a ≥ r b . By (5) this operation is associative. Also [0 −
0] is a unit element and [ a − b ] is the inverse of[ b − a ], indeed [ b − a ] + [ a − b ] = [ b − b ] = [0 − a group G .One may identify an element c ∈ M with [ c − c −
0] + [ c ′ −
0] = [( c + c ′ ) − (0 + c ′ )] + [ c ′ −
0] = [( c + c ′ ) − M is a submonoid of the group G .And conversely, given a submonoid M of a group G , for the right divisibility preorder M to be a total ordering one obviously needs(8) M ∩ ( − M ) = { } , ( − M ) + M ⊂ M ∩ ( − M ) . A Really Noncommutative Example.
Start with a kind of
Heisenberg group H overthe integers . Namely, H , as a set, is Z , its elements written as ( m, n, p ), m.n.p ∈ Z , equiv-alently as mi + nj + pk ( i, j, k being a basis). But in the operation one introduces a ‘twist’: k commutes with everything, but while i + j is as usual, j + i is defined as i + j + k , thatmaking mi + nj be as usual but nj + mi := mi + nj + ( mn ) k .In H we take a lexicographic ordering (and M will be the set of elements ≥ , , mi + nj + pk ≤ m ′ i + n ′ j + p ′ k : ⇔ m < m ′ or ( m = m ′ and n < n ′ ) or ( m = m ′ and n = n ′ and p ≤ p ′ )So M , the set of elements ≥ , ,
0) will be(10) M = (cid:8) mi + nj + pk (cid:12)(cid:12) m > m = 0 and n >
0) or ( m = n = 0 and p ≥ (cid:9) And it is clearly stable with respect to + – the ‘twist’ may affect only p , the coefficient of k ,and that only if there is an i term, but then m > p does not matter as per (10). M is not commutative: i and j belong to M and j + i = i + j + k = i + j . We claim thatstill the right divisibility ordering there is a total ordering. Indeed, Proposition 4.
For mi + nj + pk, m ′ i + n ′ j + p ′ k ∈ M , mi + nj + pk ≤ r m ′ i + n ′ j + p ′ k (right divisibility ordering) if and only if mi + nj + pk ≤ m ′ i + n ′ j + p ′ k in the lexicographicordering (9). Proof.
For a = mi + nj + pk ≤ r b = m ′ i + n ′ j + p ′ k , there must be an m ′′ i + n ′′ j + p ′′ k ∈ M ,(i.e. either m ′′ > m ′′ = 0 , n ′′ > m ′′ = n ′′ = 0 , p ′′ ≥
0) such that b = m ′ i + n ′ j + p ′ k = a + m ′′ i + n ′′ j + p ′′ k = mi + nj + pk + m ′′ i + n ′′ j + p ′′ k = ( m + m ′′ ) i + ( n + n ′′ ) j + ( p + p ′′ + nm ′′ ) k. If m < m ′ or m = m ′ and n < n ′ , i.e. the terms with i and j are not the same in a and b , thenthe term with k is of no concern and clearly the assertion holds. If these are the same, then m ′′ = n ′′ = 0 and the ‘twist’ vanishes – a and b have k terms pk and ( p + p ′′ ) k and p ′′ ≥ (cid:3) Ordering Compatible on the Right and on the Left
From now on we insist on compatibility – (C) satisfied – on the right and on the left.Cancelation is not postulated.
POSITIVE) TOTALLY ORDERED NONCOMMUTATIVE MONOIDS 5
Using The Free Noncommutative Monoid.
Recall, that the free noncommuta-tive monoid F S over a set S (referred to as alphabet and its members as letters ) is the setof words over the alphabet S with the operation of concatenation (including the empty wordwhich serves as the unit element). Here it is written multiplicatively. F S satisfies the usual universal property : Denote by i : S → F S the mapping sending eachletter s ∈ S to the one-letter word s . We say that by i , ‘ S is a subset of’ F S . Then for anymonoid M and any mapping f : S → M , f has a unique ‘extension’ F f : F S → M , i.e. suchthat F f ◦ i = f .Now endow F S with the following ordering (cid:22) : For words v, w ∈ F S , v (cid:22) w if v obtainsfrom w by deleting some of the letters, keeping the order of the letters not deleted. Call F S with this ordering the free positive (partially) ordered monoid over S .Then one easily sees that (O), (P), (C) and (A) are satisfied – F S becomes a positivepartially ordered monoid, but in general not total – (T) is not satisfied. Also if M is apositive preordered monoid, then for any f as above F f will be order-preserving.Therefore, assuming M just a monoid and ≤ a relation making it into a positive preorderedmonoid, then for the alphabet M and the identity map Id : M → M , (which, by the way,would imply the analogous facts for any alphabet S and f : S → M ), ≤ will be an extensionof the push by F Id of the ordering in F M , i.e. of the relation R := n a, b ∈ M × M (cid:12)(cid:12)(cid:12) ∃ v, w ∈ F M ( v (cid:22) w, a = F Id ( v ) , b = F Id ( w )) o , hence also will be an extension of the transitive closure of R , i.e., n a, b ∈ M × M (cid:12)(cid:12)(cid:12) ∃ n and c , c . . . , c n ( a = c , c R c , c R c , . . . , c n − R c n , b = c n ) o . That transitive closure, a relation defined canonically in any monoid M , which we denote by (cid:22) min , always satisfies (O) of course, and also, as easily seen, (P) and (C), hence makes M intoa positive preordered monoid. And we saw that any relation so making M is an extension of (cid:22) min . But if the latter satisfies (A) – is antisymmetric then so must be (cid:22) min . Thus, Proposition 5.
If in a monoid M there exists any relation making it into a positive partiallyordered monoid, then also (cid:22) min in M must satisfy (A).And if so, M with (cid:22) min is a positive partially ordered monoid. (cid:3) And M can become a positive totally ordered monoid only by some extension of (cid:22) min thatwould satisfy (O), (C), (T), (A) ( (P) always ‘inherited’ from (cid:22) min ). Remark 1.
If the monoid M we started with was commutative , then one easily finds that therelation R is just ∃ c ( b = a + c ) which is transitive hence identical with its transitive closure (cid:22) min . That is, of course, the usual way to try to define a positive ordering in a commutativemonoid.3.2. The Axiom Partially Ordered Monoids Must Satisfy.
Now, to be more explicitabout (cid:22) min , define:
Definition 6.
Let M be a (possibly noncommutative) monoid. A finite sequence of elementsin M will be called a vector (being just a word in F M as above), with the number of elementsits length and the sum of its elements (in the given order) its weight (being just the above F Id applied to it). Two vectors with the same weight will be called isobaric . A finite sequenceof vectors (in general of different lengths) will be called a table , with the vectors its rows .An augmentation of a vector v = ( a , a , . . . , a n ) is a vector obtained from v by insertingsome added element c : v ′ = ( a , a , . . . , a k , c, a k +1 , . . . , a n ) , k = 0 , , . . . , n . ELIAHU LEVY
A table A will be called monotone if each row in A is isobaric with an augmentation ofthe previous row. Now, as one easily finds,
Proposition 7.
For elements a, b in a (possibly noncommutative) monoid M , a (cid:22) min b if andonly if there is a monotone table A , such that the weight of its first row (resp. the last row) is a (resp. b ). (cid:3) As for the requirement in Prop. 5 that (cid:22) min in M satisfy (A), it says that a R a , a R a , . . . , a n − R a n , a n R a ⇒ a = a = · · · = a n , that is, that the following axiom holds: Axiom 1.
If a table A is monotone and its the first and last row are isobaric – have the sameweight, then the weights of all the rows of A are the same. If this axiom holds for a (possibly noncommutative) monoid M we shall say that the M is(positively) orderable . As wee saw, then M with (cid:22) min is a positive partially ordered monoid.From now on we always assume our monoids are orderable unless specified otherwise. Ofcourse, as we have seen, when we speak about a partially ordered monoid , in particular a totally ordered one, it is automatically orderable. i.e. Axiom 1 holds.3.3. Multiplying by a Nonnegative Integer.
Let M be a (positive) partially ordered(possibly noncommutative) monoid.In M , as in any monoid, one can multiply a natural number n by an a ∈ M : Definition 8.
For a ∈ M and n = 1 , , . . . define: na := a + . . . + a ( n times a ). Define also a = 0 . By associativity the definition is unequivocal, and we have(11) ( n + m ) a = na + ma, ( mn ) a = m ( na ) . Also, by Def. 8 m ≤ n ⇒ ma ≤ na , and by (P) and (C) a ≤ b ⇒ na ≤ nb . Corollary 9. (to Axiom 1) For a ∈ M and n < n < n , if n a = n a then they are equalalso to n a . Proof .
Take in Axiom 1 a table A with n − n + 1 rows all whose entries are a , the i ’throw of length n + i − A is obviously monotone. Since n a = n a its first and last row areisobaric, thus by Axiom 1 the weights of all the rows are the same, hence the assertion. (cid:3) Absorbing Elements and Idempotents.Definition 10. a absorbs b (and b is absorbed by a ) on the left if a + b = a .Similarly, a absorbs b (and b is absorbed by a ) on the right if b + a = a . a is an idempotent if a + a = a (that is, a absorbs itself on the left/right). a is a generalized idempotent if na = ma for some natural numbers n = m . Thus, if a is not a generalized idempotent, than n < m ⇒ na < ma .Clearly, if pa is a generalized idempotent for some p = 1 , , . . . then so is a . Remark 2.
Suppose a is a generalized idempotent. Then ∃ n < m na = ma and we take thepair n, m with the least possible n . This means that a, a, . . . , na are all different.Then by Cor. 9 na, ( n + 1) a, . . . , ma are all equal, and by (11) also ( n + k ) a, . . . , ( m + k ) a are equal for k = 1 , , . . . . And since for k and k + 1 these sequences overlap, we finally findthat na = ( n + 1) a = ( n + 2) a = . . . . POSITIVE) TOTALLY ORDERED NONCOMMUTATIVE MONOIDS 7
Definition 11.
Let P be a partially ordered set.A subset S ⊂ P is called lower , (resp. upper ) hereditary if a ∈ S, b ≤ a ⇒ b ∈ S (resp. a ∈ S, b ≥ a ⇒ b ∈ S ).For a ∈ S , the lower cone , (resp. upper cone ) with vertex a is the set (cid:8) b ∈ S (cid:12)(cid:12) b ≤ a (cid:9) , (resp. (cid:8) b ∈ S (cid:12)(cid:12) b ≥ a (cid:9) ). In particular, in a (positive) partially ordered (possibly noncommutative) monoid M wemay speak of lower hereditary submonoids . Then for an element a ∈ M , the lower hereditarysubmonoid generated by a is easily seen to be(12) (cid:8) b ∈ M (cid:12)(cid:12) ∃ n = 1 , , . . . ( b ≤ na ) (cid:9) . Also, the sets of elements absorbed on the left (resp. on the right) by a fixed element a in M ,lAb( a ) := (cid:8) b ∈ M (cid:12)(cid:12) a + b = a (cid:9) , rAb( a ) := (cid:8) b ∈ M (cid:12)(cid:12) b + a = a (cid:9) , are clearly submonoids and by properties (P) (C) and (A) of the ordering also lower hereditary.Since one always has b ≤ a + b, b + a , we conclude that if b is absorbed on the left or on theright by a then b ≤ a .So lAb( a ) and rAb( a ) are lower hereditary sumonoids which are contained in the lower conewith vertex a – the latter, of course, lower hereditary but need not be a submonoid (i.e. neednot contain the sum of two of its elements).And we clearly have Proposition 12.
Let M be a (positive) partially ordered (possibly noncommutative) monoid.Then for some fixed a ∈ M the following are equivalent: (1) lAb ( a ) is the whole lower cone with vertex a , that is, a absorbs on the left all theelements ≤ from it. (2) rAb ( a ) is the whole lower cone with vertex a , that is, a absorbs on the right all theelements ≤ from it. (3) a is an idempotent: a + a = a , i.e. a belongs to lAb ( a ) , equivalently belongs to rAb ( a ) . (4) The lower cone with vertex a is a (lower hereditary) submonoid. (5) lAb ( a ) has a greatest element, i.e. ≥ than all its other elements (necessarily equal to a ). (6) rAb ( a ) has a greatest element, i.e. ≥ than all its other elements (necessarily equal to a ). (cid:3) Remark 3.
To see that there are positive partially ordered (possibly) noncommutativemonoids where some implications will not hold, say where for some a and b , a + b is a gener-alized idempotent while a and b are not, consider the following construction:Start with the free noncommutative momoid F S over an alphabet S , partially ordered by (cid:22) as above. Let C be some upper hereditary subset of F S , say the upper cone of some w ∈ F S .Now take the equivalence relation ≈ in F S just collapsing C to a point [ C ]. Note that then F S / ≈ will always be a (positive partially ordered) monoid – the monoid operation there well-defined, making [ C ] a greatest element there (hence absorbing all others on the left and onthe right) and with no relations between words on S except the equality among all members of C .For example: if a, b ∈ S and C is the upper cone of the word ( a, b ), then a + b is anidempotent in F S / ≈ while neither of a , b is a generalized idempotent! ELIAHU LEVY
On the other hand one notes that since n ( b + a ) = b + ( n − a + b ) + a we do have always: for a and b in a monoid M , b + a is a generalized idempotent if and only if a + b is so .One may loosely say that when multiplying by a large natural number, the difference between a + b and b + a somewhat dwindles . We shall pursue this clue in § The Case of Total Ordering, Ratios
Let M be a (positive) totally ordered (possibly noncommutative) monoid.Then a lower (resp. upper) hereditary subset is a lower ray ( upper ray ) (possibly empty orthe whole M ) and a set is a lower ray if and only if its complement is an upper ray.4.1. Archimedicity and Possible Absorbing.Definition 13.
For a, b ∈ M , a is said to be infinitely greater than b , (and b infinitelysmaller than b ) if there is no n = 1 , , . . . such that a ≤ nb , in other words, a > nb for all n . Otherwise, i.e. if there exists an n with a ≤ nb , then a is called Archimedean to b .If a and b are both Archimedean to each other, we say that a and b are commensurable . Remark 4.
Clearly the following relations are transitive: being infinitely greater than; beingArchimedean to; being commensurable to. Also, for a fixed a ∈ M , M is partitioned into threesets (possibly empty): the lower ray of the b ’s infinitely smaller then a ;, all these smaller thenthe b ’s commensurable to a ; – these two sets together make those to which a is Archimedean;,and all these smaller than the b ’s infinitely greater than a which make an upper ray.And, (cf. the previous § ), for an a ∈ M , the lower hereditary submonoid generated by a is (cid:8) b ∈ M (cid:12)(cid:12) ∃ n = 1 , , . . . ( b ≤ na ) (cid:9) , thus it coincides with the set of elements of M which are Archimedean to a .In particular one notes that the set of elements Archimedean to a is a (lower hereditary)submonoid – contains the sum of any two of its members.Then we have: Proposition 14.
For a, b ∈ M , a absorbing b on the left or right is possible only if either a is an idempotent: a = a + a (hence absorbs all the elements ≤ from it), or a is infinitely greater than b . Proof .
By the previous § , the set of elements absorbed by a on the left (resp. on the right)(which is contained in the lower cone with vertex a ) is a lower hereditary monoid, hencecontains the lower hereditary monoid generated by b which is the set of elements Archimedeanto b .Then either the latter is the whole lower cone with vertex a , equivalently contains a : a isan idempotent.Or it does not contain a , thus a is not Archimedean to b : a is infinitely greater than b . (cid:3) So Proposition 15.
For commensurable a, b ∈ M , a absorbing b on the left or right is possibleonly if a = nb for some natural n , so we are in the situation of Remark 2, in particular b isa generalized idempotent. Proof.
By the previous proposition, a must be an idempotent. Also since a is also Archimedeanto b , there is a natural number n such that a ≤ nb . Let n be the smallest satisfying that. If n = 1 than b ≤ a ≤ b ⇒ a = b . If n > b ≤ a , so a absorbs b hence also absorbs nb thusone has a ≥ nb and finally a = nb . (cid:3) POSITIVE) TOTALLY ORDERED NONCOMMUTATIVE MONOIDS 9
And moreover
Proposition 16. If a and b are commensurable, then if one of them is a generalized idempotentso is the other, and in fact there are natural numbers n and m such that a, a, . . . , na aredifferent, b, b, . . . , nb are different, while na is equal to mb and na = ( n + 1) a = ( n + 2) a = . . . = mb = ( m + 1) b = ( m + 2) b = . . . .And the latter element, an idempotent, is the greatest in the lower ray of all elements towhich a (equivalently b ) is Archimedean, and absorbs them on the right and on the left. Proof.
Suppose a is a generalized idempotent and n as in Remark 2. Then the ‘maximal’ na is an idempotent, absorbing itself hence elements smaller than it on the left and on the right.Now b is commensurable with a . Therefore there is some n ′ so that b ≤ n ′ a , thus b , henceall its multiples, are less or equal than the maximal idempotent na . On the other hand a ,hence na , is commensurable with b . So there is an m such that na ≤ mb , take the minimalwith this property. But then mb = na and the assertion follows. (cid:3) Ratios `a la Euclid’s
Elements . Now let us try to follow, in our setting, what Euclid’s
Elements does. There, in order to impose numerical values on geometrical objects which, atthe start, one just adds and compares, a theory of ‘ratios’ (due to Eudoxos) is developed, amost abstract and almost ‘modern’ by the standards of the
Elements . (Where they alreadyknew that they cannot expect all ratios to be ‘rational’ – given by a fraction of two integers– the Pythagoreans had discovered, two centuries before, that the ratio of the diagonal of asquare to its side is irrational!)So, let a, b ∈ M , commensurable and not generalized idempotents. Then by Prop. 15 noneof them can absorb the other.Let m, n be positive integers. We put a sign = , ≥ , ≤ , >, < between the ratio a : b and theratio m : n if that sign holds between na and mb .Now, if a : b = m : n , i.e., na = mb then we have for any positive integer p , ( pn ) a = p ( na ) = p ( mb ) = ( pm ) b , meaning a : b = ( pm ) : ( pn ).And if a : b ≤ m : n , i.e., na ≤ mb , then ( pn ) a = p ( na ) ≤ p ( mb ) = ( pm ) b , i.e., a : b ≤ ( pm ) : ( pn ).This means that for a fraction q = m/n , the signs a : b has with two ways to write it as aratio between a numerator and a denominator, namely ( p m ) : ( p n ) and ( p m ) : ( p n ), maycertainly not be < vs. > or > vs. < – they are both ≤ or = (write then a : b ≤ q ) or both ≥ or = (writing a : b ≥ q ).Now suppose the fractions q ′ < q . Write them with a common denominator as q ′ = m ′ /n, q = m/n, m ′ < m . Then if a : b ≤ q ′ , i.e., na ≤ m ′ b , then, since m ′ b < mb , ( b is nota generalized idempotent!), we have na < mb , i.e., a : b < m : n , so a : b ≥ q does not hold .Similarly if a : b ≥ q then we find a : b > m ′ : n hence a : b ≤ q ′ does not hold.So we conclude that, given a, b ∈ M , we are in the situation featuring with the Elements ’‘ratios’. Namely, the set of all positive rational numbers q is divided into two sets: those where a : b ≥ q and those where a : b ≤ q , and these form a Dedekind cut – both are nonempty since a and b are commensurable. The former is of the form [ α, → ) or ( α, → ) and the latter is ofthe form (0 , α ] or (0 , α ), for some real α > α as the ratio a : b . Thus a : b ∈ (0 , + ∞ ). Properties of Ratios.Proposition 17.
For a, b, c ∈ M commensurable, not generalized idempotents, (1) a : a = 1 . (2) ( b + a ) : ( a + b ) = 1(3) ( na ) : a = n (4) a : c = ( a : b ) · ( b : c ) . (5) b : a = ( a : b ) − . (6) a ≤ b ⇒ a : c ≤ b : c and c : a ≥ c : b . (7) ( a + b ) : c = ( b + a ) : c = ( a : c ) + ( b : c ) . (8) If a and b are commensurable and not generalized idempotents, but c is infinitelysmaller than a (equivalently than b ), then ( c + a ) : b = ( a + c ) : b = a : b , a :( c + b ) = a : ( b + c ) = a : b . (9) In particular, also in the case that c is infinitely smaller than a we have ( a + c ) :( c + a ) = 1 . (10) Conversely, for b Archimedially related to a , thus not infinitely smaller than it, suchequalities cannot occur. Indeed then ( a + c ) : a > , ( c + a ) : a > , a : ( a + c ) < and a : ( c + a ) < . Proof . (1) Obviously, if n ≤ m then na ≤ ma . Hence n : m ≤ a : a and m : n ≥ a : a .Therefore n/m > a : a cannot hold and m/n < a : a cannot hold. Having thatwhenever n ≤ m , a : a must be 1.(2) We have ( n + 1)( a + b ) = a + n ( b + a ) + b ≥ n ( b + a ). Thus ( n + 1) : n ≥ ( b + a ) : ( a + b )and similarly ( n + 1) : n ≥ ( a + b ) : ( b + a ), i.e. n : ( n + 1) ≤ ( b + a ) : ( a + b ). Therefore( b + a ) : ( a + b ) cannot be > ( n + 1) /n and cannot be < n/ ( n + 1). That being thecase for every n , ( b + a ) : ( a + b ) must be 1.(3) Suppose ( na ) : c < n (resp. ( na ) : a > n ). Then there must be natural numbers m, p such that ( na ) : a > p/m, p/m > n (resp. ( na ) : a < p/m, p/m < n ). This meansthat p > mn while ( mn ) a ≤ pa is impossible (resp. p < mn while ( mn ) a ≥ pa isimpossible). But, p > mn surely implies ( mn ) a ≤ pa (resp. p < mn surely implies( mn ) a ≥ pa ) – contradiction.(4) Suppose a : c < ( a : b ) · ( b : c ) (resp. a : c < ( a : b ) · ( b : c )). Then there mustbe natural numbers n, m, p such that a : c > p/n, a : b < m/n, b : c < p/m (resp. a : c < p/n, a : b > m/n, b : c > p/m ). This means that na ≤ pc is impossibleand so are na ≥ mb and mb ≥ pc (resp. na ≥ pc is impossible and so are na ≤ mb and mb ≤ pc ). But, the ordering being total, that means na > pc > mb > na (resp. na < pc < mb < na ) – impossible.(5) The assertion follows, of course, from items 4 and 1, but one may give a direct proof:It is enough to show that for any n, m , n/m less than (resp. greater than) b : a implies the same (even if in the weak sense) with ( a : b ) − . But n/m < b : a ⇒ n : m cannot be ≥ b : a ⇒ na ≥ mb does not hold ⇒ m : n ≤ a : b does not hold ⇒ m/n ≥ a : b ⇒ n/m ≤ ( a : b ) − . And similarly for the reverse inequalities.(6) The second assertion follows from the first and item 5. For the first assertion, we haveto show that n/m less than a : c implies the same (even if in the weak sense) with b : c . Indeed n/m < a : c ⇒ n : m cannot be ≥ a : c ⇒ nc ≥ ma does not hold ⇒ a fortiori , as a ≤ b hence ma ≤ mb, nc ≥ mb does not hold ⇒ n : m ≥ b : c does not hold ⇒ n/m ≤ b : c. POSITIVE) TOTALLY ORDERED NONCOMMUTATIVE MONOIDS 11 (7) Here, in order to partially circumvent the noncommutativity, we rely on the results ofthe previous items. Take c fixed and consider the map ρ c : a a : c between the set ofelements in M commensurable to c and (0 , ∞ ) ⊂ R , the latter endowed with addition(of numbers). We wish to prove that it preserves the operation.By the previous items we know that, for variable a, b ∈ M , ρ c ( c ) = 1, a : b = ρ c ( a ) /ρ c ( b ), and a ≤ b ⇒ ρ c ( a ) ≤ ρ c ( b ) – ρ c is monotone. Also by item 3 ρ ( na ) = nρ ( a ). And since ( b + a ) : ( a + b ) = 1 one obtains that ρ c ( b + a ) = ρ c ( a + b ).Thus the value ρ c gives to a sum does not depend on the order of the summands.In particular nρ c ( a + b ) = ρ c ( n ( a + b )) = ρ c ( na + nb )So suppose that for some a and b ( a + b ) : c < ( a : c ) + ( b : c ) (resp. ( a + b ) : c > ( a : c )+( b : c )). Then there must be natural numbers n, m, p such that a : c < m/n, b : c
( m + p ) /n (resp. a : c > m/n, b : c > p/n, ( a + b ) : c < ( m + p ) /n ).This means that n ( a + b ) ≤ ( m + p ) c is impossible and so are na ≥ mc and nb ≥ pc (resp. n ( a + b ) ≥ ( m + p ) c is impossible and so are na ≤ mc and nb ≤ pc ). The ordering beingtotal, that means na + nb ≤ ( m + p ) c < n ( a + b ) (resp. na + nb ≥ ( m + p ) c > n ( a + b )).Applying ρ c which is monotone and gives to both na + nb and n ( a + b ) the same value nρ c ( a + b ) and to ( m + p ) c the value m + p , we find ( a + b ) : c = ρ c ( a + b ) = ( m + p ) /n .That contradicting the strict inequality here that we had.(8) First, since c is infinitely smaller than a , implying of course c < a , and also a + c ≥ c , a + c and a are surely commensurable. By items 4 and 5 it suffices to prove ourassertions for some fixed b .Thus let a and c vary and use the notation of item 7, for some fixed a : ρ a ( a ) := a : a and also β ( a, c ) := ρ a ( a + c ) − ρ a ( a ), β ′ ( a, c ) := ρ a ( c + a ) − ρ a ( a ). Then, since a + c, c + a ≥ a , β ( a, c ) , β ′ ( a, c ) ≥ ρ described in 7.First they yield that β = β ′ , both not depending on a . Indeed, for any a and b , β ( a, c ) := ρ a ( a + c ) − ρ a ( a ) = ρ a ( a + c + b ) − ρ a ( b ) − ρ a ( a ) = ρ a ( c + b ) − ρ a ( b ) = β ′ ( b, c ) . Thus β ( a, c ) = β ′ ( a, c ) = β ( c ).Now this β is clearly additive in c , since β ( c + c ′ ) = ρ a ( a + c + c ′ ) − ρ a ( a ) = ( ρ a ( a + c + c ′ ) − ρ a ( a + c ))+( ρ a ( a + c ) − ρ a ( a )) = β ( c ′ )+ β ( c ) . In particular β ( nc ) = nβ ( c ).But for any a and b , commensurable while c infinitely smaller than them, c is surely smaller than b . Therefore a + c ≤ a + b making β ( c ) ≤ ρ a ( a + b ) − ρ a ( a ) = ρ a ( b ) , – any value taken by ρ a at any b bounding β ( c ) for any c ! But then it also bounds β ( nc ) = nβ ( c ) for any n . So β must vanish.(9) Follows from the previous item.(10) That follows from item 7 and from c : a being always positive . (cid:3) So one may loosely say that for any (positive) totally ordered monoid, as far as commensu-rable elements, not generalized idempotents, are concerned, and these up to an ‘error’ infinitelysmaller than them, then the ‘ratio’, as a ‘first approximation’, imposes a numerical additivestructure, thus commutative . Acknowledgement.
I am much grateful to Professor Orr Shalit for posing and remarkingon the problem which is the impetus for this note and on this note in preparation. Some ideasin this note are from him.
References [1] A. H. Clifford and G. B. Preston,
The algebraic theory of semigroups I , Mathematical Surveys, no. 7,American Mathematical Society, 1961.[2] I. N. Herstein,
Noncommutative Rings , The Carus Mathematical Monographs, no. 15, The MathematicalAssociation of America, 1968.[3] Orr Moshe Shalit and Michael Skeide,
CP-Semigroups and Dilations, Subproduct Systems and Superprod-uct Systems: The Multi-Parameter Case and Beyond
March 2020, arXiv:2003.05166v1
Department of Mathematics, Technion – Israel Institute of Technology, Technion City,Haifa 3200003, Israel
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