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Publicly Available Published by De Gruyter September 20, 2016

The structure group of an L-algebra is torsion-free

  • Wolfgang Rump EMAIL logo
From the journal Journal of Group Theory

Abstract

L-algebras arise in algebraic logic, in the theory of one-sided lattice-ordered groups, and in connection with set-theoretic solutions of the quantum Yang–Baxter equation. They apply in several ways to Garside groups. For example, the set of primitive elements, the set of simple elements, and the negative cone of a Garside group are all L-algebras. Picantin’s iterated crossed product decomposition of Garside groups can be reformulated and extended in terms of L-algebras. It is proved that the structure group of an L-algebra, introduced in connection with the “logic” of -groups, is torsion-free. This applies to the left group of fractions of not necessarily noetherian, Garside-like monoids which need not embed into their ambient group.

A topological proof that Artin’s braid groups are torsion-free was given in 1962 by Fadell, Fox, and Neuwirth [12, 13]. The first algebraic proof made use of the orderability of braid groups [5, 16]. Later, Dehornoy [6] showed that all Garside groups are torsion-free, and gave a much simpler proof [8] without the noetherian hypothesis. The method of simplifying proofs by dropping redundant hypotheses came to a certain end with a one-line proof that every group with a right invariant lattice order is torsion-free (see [20, Proposition 3]). On the other hand, Dehornoy’s slightly longer proof [8] applies to an even more general situation. In the terminology of Section 1, his result shows that the group of fractions of a cancellative left hoop (a residuated monoid) is torsion-free.

In this paper, we prove that left cancellability can be dropped. More generally, we replace the left hoop by a unital cycloid [18], a set with a single binary operation which satisfies

(0.1)(xy)(xz)=(yx)(yz).

The relevance of equation (0.1) will be explained in Section 1. We give an example where the natural map from the unital cycloid to its structure group – the “group of fractions” – is surjective. So our theorem does not rely on any kind of partial ordering of the group.

1 The structure group of an L-algebra

Equation (0.1) exhibits a close relationship between group theory, algebraic logic, and quantum structures [11]. It first occurred in a logical context [2, 21]. Here xy stands for the logical implication (“x implies y”). If the left multiplications yxy are bijective, such a set (X;) is called a cycle set [17]. By [17, Proposition 1], cycle sets are in one-to-one correspondence with non-degenerate unitary set-theoretic solutions of the quantum Yang–Baxter equation. Thirdly, equation (0.1) occurs in the theory of lattice-ordered groups (-groups for short) [1, 4], the negative cone G-:={xG:x1} of an -group G being a cycloid with respect to

xy:=(yx-11).

As a fourth instance, every Garside monoid satisfies equation (0.1) (see [7, equation (1.5)]).

In the latter two examples, there is an associative multiplication which should be distinguished from the operation xy. Therefore, we write xy instead of xy, which reflects the logical origin of that operation. In the presence of a multiplication, we then have

(1.1)xyzxyz.

By this equivalence, the unit element of an -group G satisfies

(1.2)xx=x1=1,1x=x

for all xG-. A cycloid (X;) with such an element 1 (a logical unit [18]) is said to be unital. (In propositional logic, 1 stands for the “true” proposition.)

For a unital cycloid (X;), the relation

xy:xy=yx=1

is a congruence. If this relation is trivial, that is, xy implies x=y, then X is called an L-algebra [18]. Thus every unital cycloid X gives rise to an L-algebra X/. Every L-algebra has a partial order

xy:xy=1

similar to the logical entailment relation.

The negative cone, and every interval [a,1] of an -group is an L-algebra in two ways, namely, with respect to each of the operations xy:=yx-11 and xy:=x-1y1. In several respects, L-algebras play a fundamental part in the theory of Garside groups. For example, the primitive elements as well as the simple elements [10, 7] of a Garside group form an L-algebra. Picantin’s iterated crossed products of Garside groups [15] admit a simple formulation in terms of L-algebras [19]. More generally, L-algebras arise in connection with right -groups, that is, groups with a right invariant partial order. For example, the structure groups of non-degenerate unitary set-theoretic solutions X×XX×X of the quantum Yang–Baxter equation are right -groups with a distributive lattice structure [20]. For finite X, these groups are Garside [3, 9].

In [18] we associate a structure groupG(X) to any L-algebra X. To explain this, we have to clarify the rôle of the product in (1.1). Define a left hoop [18] to be a monoid H with a binary operation satisfying

aa=1,
abc=a(bc),
(ab)a=(ba)b

for a,b,cH. By [18, Propositions 3 and 4], every left hoop is an L-algebra, and any pair a,bH has a meet

(1.3)ab=(ab)a.

Therefore, the partial order of H is given by

(1.4)abcH:a=cb.

A left hoop H is self-similar if it is right cancellative. By [18, Proposition 5], this can be expressed in terms of equations:

Proposition 1

For a left hoop H, the following are equivalent.

  1. H is self-similar, that is, ac=bca=b holds for a,b,cH.

  2. aba=b for all a,bH.

  3. abc=((ca)b)(ac) for all a,b,cH.

Since aa=1 follows by (b), a self-similar left hoop is given by the axioms

aba=b,
abc=a(bc),
(ab)a=(ba)b.

Note that by equation (1.1), self-similarity is a property of L-algebras. In fact, an L-algebra X is self-similar if and only if for any xX, the map y(xy) gives a bijection {yX:yx}X (see [18, Definition 2]). Moreover, self-similarity implies that X admits a unique multiplication which makes X into a left hoop.

Now let H be a self-similar L-algebra with an L-subalgebraX, which means that the operation of X is induced by that of H. As remarked above, H can be regarded as a left hoop. If the monoid H is generated by X, we call H a self-similar closure of X. By [18, Theorem 3], a self-similar closure of any L-algebra X exists and is unique, up to isomorphism. We denote it by S(X). In particular, every self-similar left hoop H satisfies

S(H)=H.

Since any self-similar left hoop satisfies the left Ore condition, the self-similar closure S(X) of an L-algebra X admits a left group of fractions G(X) with a natural map

(1.5)q:XS(X)G(X).
Example 1

Classical logic is built upon the two-element L-algebra Xa:={1,a}, which is completely determined by the property of 1 being a logical unit. Thus a<1, where a and 1 stand for the two possible truth values. The self-similar closure S(Xa) is the negative cone of the -group G(Xa), the additive group of integers. Thus S(Xa):=G(Xa)- consists of the powers 1>a>a2>a3> with aman=an-m for mn. As the two-element L-algebra Xa generates the group G(Xa), this shows that logic creates arithmetic, the theory of the ring End(G(Xa)).

Example 2

If the pair of truth values is replaced by the interval [-1,0], we obtain an L-algebra with xy:=min{x,y}-x, which is fundamental for measure and integration theory [14]. Its self-similar closure is the negative cone of G([-1,0]), the additive group of the reals.

Example 3

Let V be an (n+1)-dimensional Euclidean space, and let (V) be the corresponding elliptic real n-space, that is, the projective space 𝒫(V) of one-dimensional subspaces of V. The scalar product of V defines an elliptic polarity 𝒫(V)𝒫(V) which associates a hyperplane P to any point P𝒫(V). “Elliptic” means that PP for all P𝒫(V). Let (V) denote the lattice of -linear subspaces of V. Thus U𝒫(U) gives a lattice isomorphism between the U(V) and the projective subspaces of (V). Define

HL:=(HL)H

for H,L(V). Then 1:=(V) is a logical unit, and

HL(HL)H=1H=((HL)H)H=(HL)(HH)=HL,

which gives

HLHL=1.

Furthermore, ((HL)H)H=(HL)(HH)=HL, which yields

(HL)H=HL.

Note that HHL. For H,L,M(V), this implies that

(HL)(HM)=(H(HL))(HM)
=H(HL(HM))
=(HLM)H.

Hence

(HL)(HM)=((HL)(HM))(HL)=(HLM)H(H(HL))=(HLM)((HH)(HL))=(HLM)(HL)=(HL)M,

which shows that (V) is an L-algebra. Assume that H and L are orthogonal (HL) in the sense that HL or equivalently, LH. Then

HL=H(HL)=(HH)L=L.

So L(HM)=(HL)(HM)=(HL)M, which yields

HLMLHM.

Thus HL coincides with the product HLS((V)) whenever HL. Let X(V)(V) be the L-subalgebra of subspaces HV of codimension 1. Since every H(V) is a finite meet of pairwise orthogonal hyperplanes, it follows that S((V))=S(X(V)). By [20, Proposition 19], S(X(V)) is a modular lattice.

Furthermore, it can be shown that the equation

(HL)(LH)=H

holds for distinct H,LX(V). By [20, Theorem 4], this implies that G(X(V)) is a Garside group. The example gives an L-algebraic interpretation of the well-known Gram–Schmidt process in the Euclidean space V.

For a unital cycloid X, we define the structure groupG(X) to be G(X/). Note that in general, the map (1.5) need not be injective. Indeed, two elements x,y of an L-algebra X satisfy q(x)=q(y) if and only if cx=cy holds for some cS(X). Thus q is injective if and only if S(X) is left cancellative. Two elements x,yX with q(x)=q(y) are called equipollent [18]. If X has a smallest element 0 (= “false”), equipollence of x and y is equivalent to x0=y0. Our aim is to show that the structure group of any L-algebra X is torsion-free. Thus, by (1.5), we can assume, without loss of generality, that X is a self-similar left hoop.

Let H be a self-similar left hoop. For a pair of elements a,bH, we define the derived pair(a,b)H2 by

(1.6)a:=ba,b:=ab.

The n-th derived pair will be denoted by (a(n),b(n)) for n. In particular, a(0):=a and b(0):=b. The notation does not express the dependency of a(n) on both a and b, which should always be clear from the context. Furthermore, we write

(1.7)an:=a(n-1)aa,bn:=b(n-1)bb

for n. Thus a0=b0=1. More generally, we abbreviate

(1.8)am,n:=a(m-1)a(n),bm,n:=b(m-1)b(n)

for mn in . The concept of derived pair (1.6) is justified by

Proposition 2

Let H be a self-similar left hoop. Two elements a,bH are equipollent if and only if a and b are equipollent.

Proof.

By Proposition 1, we have

acb=((ba)c)(ab)=(ac)b.

Hence

cacb=c(ac)b=((bc)(ac))(cb).

So cacb is equivalent to the conjunction of bcac and cb. Thus a and b are equipollent if and only if ac=bc for some ca,b. By [18, Proposition 12], the latter condition is equivalent to q(a)=q(b).∎

The product expressions (1.7) satisfy the following.

Lemma 1

Let H be a self-similar left hoop. For a,bH and m,nN,

(1.9)a(m+n)=(ambn)a(m),b(m+n)=(bnam)b(n).

Proof.

We proceed by induction over the sum s:=m+n. For s=0 or n=0, the equations are obvious. Assume that the equations hold for a fixed sum s. For s+1=m+(n+1) with m,n, using Proposition 1, this gives

a(m+n+1)=b(m+n)a(m+n)=((bnam)b(n))((ambn)a(m))=((bnam)b(n))(ambn)a(m)=(amb(n)bn)a(m)=(ambn+1)a(m).

By symmetry, the lemma is proved. ∎

The expressions in equation (1.8) admit the following interpretation.

Proposition 3

Let H be a self-similar left hoop. For a,bH and m,nN,

ambn=bm+n,m,bnam=am+n,n.

Proof.

We prove the first equation by induction on n. For n=0, the equation is trivial. Thus, assume that the equation holds for a fixed n. Then Lemma 1 gives

ambn+1=amb(n)bn
=((bnam)b(n))(ambn)
=b(m+n)bm+n,m=bm+n+1,m,

which completes the inductive step.

The second equation is proved analogously. ∎

The following corollary is essentially due to Dehornoy [8].

Corollary 1

Let H be a self-similar left hoop. For a,bH and m,nN,

(1.10)ambn=am+n,nbn=bm+n,mam.

Proof.

This follows immediately by equation (1.3). ∎

Corollary 2

Let H be a self-similar left hoop. For a,bH and nN, we have

(1.11)(q(a)-1q(b))n=q(an)-1q(bn).

Proof.

Equation (1.11) can be rewritten as

q(an)(q(a)-1q(b))n=q(bn).

For n=0, this is obvious. Assume the equation holds for some n. Then equation (1.10) gives

q(an+1)(q(a)-1q(b))n+1=q(an+1,1)q(b1)(q(a)-1q(b))n
=q(bn+1,n)q(an)(q(a)-1q(b))n
=q(bn+1,n)q(bn)
=q(bn+1).

So the corollary follows by induction. ∎

Definition 1

Let H be a self-similar left hoop. For a pair a,bH, we define a weak (a,b)-cycle to be a non-empty sequence (c1,,cn)Hn with

bci=aci+1for 1i<n  and  q(ac1)=q(bcn).

If ac1=bcn, we speak of an (a,b)-cycle.

An (a,b)-cycle can be visualized by a periodic diagram, e. g., for n=3:

So we can assume that (ci) is an n-periodic sequence defined for all i, and there is a single condition bci=aci+1 for all i.

Proposition 4

Let H be a self-similar left hoop. If the sequence (c1,,cn)Hn is an (a,b)-cycle, then (ac1,,acn) is an (a,b)-cycle.

Proof.

For 1i<n, equation (1.10) yields baci=abci=aaci+1. Furthermore, aac1=abcn=bacn, which proves the claim. ∎

Lemma 2

If a self-similar left hoop H admits an (a,b)-cycle, then q(a)=q(b).

Proof.

Let (c1,,cn)Hn be an (a,b)-cycle. We show that

(1.12)bicj=aici+j

holds for i and j. For i=0, this is trivial. Assume that equation (1.12) is valid for a given i. Then equation (1.10) implies that

bi+1cj=b(i)bicj
=bi+1,iaici+j
=ai+1,1b1ci+j
=ai+1,1aci+j+1
=ai+1ci+1+j,

which proves equation (1.12). In particular, bncj=ancj. Hence

bn=a1bn=an+1,nbn,

which yields a(n)=an+1,n=1. Similarly, an=anb1=bn+1,nan implies that b(n)=1. By Proposition 2, this proves that a and b are equipollent. ∎

Proposition 5

If a self-similar left hoop H admits a weak (a,b)-cycle, then q(a) is equal to q(b).

Proof.

Let (c1,,cn)Hn be a weak (a,b)-cycle. So there is an element cH with cac1=cbcn. Choose dcbcnbci for all i{1,,n}. Since dbci, there exist elements diH with d=dibci for all i{1,,n}. In particular, dnbcn=dcbcn shows that dnc. Hence we have dnac1=dnbcn. Furthermore, di+1bci+1=d=dibci=diaci+1 gives dia=di+1b for 1i<n, and d1bc1=d=dnbcn=dnac1 yields dna=d1b. So the di extend to an n-periodic sequence (di) with d0ac1=d0bcn and

dia=di+1b

for all i. Hence

(di-1di)dib=(di-1di)di-1a
=(didi-1)dia
=(didi-1)di+1b,

which yields

(1.13)(di-1di)di=(didi-1)di+1

for all i. Similarly,

(di+1di)dia=(di+1di)di+1b
=(didi+1)dib
=(didi+1)di-1a

gives

(1.14)(di+1di)di=(didi+1)di-1

for all i. Now equations (1.13) and (1.14) give

di-1di=di(didi-1)di+1=((di+1di)(didi-1))(didi+1)=((di+1di)didi-1)(didi+1)=((didi+1)di-1di-1)(didi+1)=didi+1,

which shows that the element

p:=didi+1

does not depend on i. Similarly,

di+1di=di(didi+1)di-1
=((di-1di)(didi+1))(didi-1)
=((di-1di)didi+1)(didi-1)
=((didi-1)di+1di+1)(didi-1)
=didi-1.

So the element

q:=didi-1

does not depend on i. Hence we obtain

pdi=(didi+1)di=(di+1di)di+1=qdi+1

for all i. Thus (d1,,dn) is a (q,p)-cycle. By Lemma 2, this shows that p and q are equipollent. In particular, d0d1 and d1d0 are equipollent. So Proposition 2 implies that d0 and d1 are equipollent. Since d0a=d1bab, there is an element eH with d0a=e(ab). Hence we have d0=e(ab) and d1=e(ba). Thus ab and ba are equipollent. Again by Proposition 2, this yields q(a)=q(b). ∎

Now we are ready to prove our main result.

Theorem

The structure group G(X) of an L-algebra X is torsion-free.

Proof.

Every element of G(X) is of the form q(a)-1q(b) with a,bS(X). Assume that (q(a)-1q(b))n=1 for some n2. Then equation (1.11) gives

q(an)=q(bn).

Define ci:=bn-1,n-ian-i for i{1,,n}. Then equation (1.10) gives

b(n-1)ci=bn,n-ian-i
=an,ibi
=a(n-1)an-1,ibi
=a(n-1)bn-1,n-i-1an-i-1
=a(n-1)ci+1

for 1i<n. Furthermore,

q(a(n-1)c1)=q(a(n-1)bn-1,n-1an-1)
=q(an)
=q(bn)
=q(b(n-1)bn-1,0a0)
=q(b(n-1)cn).

This shows that (c1,,cn) is a weak (a(n-1),b(n-1))-cycle in S(X). Thus a(n-1) and b(n-1) are equipollent by Proposition 5. So Proposition 2 implies that q(a)=q(b). Whence q(a)-1q(b)=1. ∎

Since every self-similar left hoop H is an L-algebra and satisfies S(H)=H, we have:

Corollary

The left group of fractions G(H) of a self-similar left hoop H is torsion-free.

2 Examples and constructions

We start with two new constructions of self-similar left hoops. Let M be a self-similar left hoop. By S(M) we denote the group of bijections

α:M{1}M{1}

with (xy)α=x(yα) for all x,yM{1}. The equivalence (1.4) shows that S(M) consists of order automorphisms, that is,

xyxαyα

holds for x,yM{1} and αS(M). Note that the equation (xy)α=x(yα) is valid also for x=1, while for y=1, it turns into xα=x(1α). So the αS(M) can be conceived as right multiplications with an “external” element 1α.

Definition 2

Let H and M be self-similar left hoops, and let σ:HS(M) be a monoid homomorphism. We write xa:=σ(a)(x). Define the lexicographic semidirect product×HM to be the set of all formal products ax with aH and xM such that

(2.1)axby:={(ab)yfor x=1,a(xby)for x1,

for a,bH and x,yM.

Lemma 3

Let H be a monoid with a partial order (1.4) and a binary operation ab which satisfies abcabc for all a,b,cH. Then H is a left hoop.

Proof.

The inequality 1aa is trivial. For all dH, we have

dabcdabcdabcda(bc).

Hence we have abc=a(bc). To verify equation (1.3), we note first that (ab)aa,b. Conversely, assume that ca,b. Then c=dab for some dH. Hence dab, and thus d=e(ab) for some eH. So we obtain c=da=e(ab)a(ab)a. ∎

Proposition 6

Let H and M be left hoops with M self-similar. The lexicographic semidirect product ×HM is a left hoop which satisfies

(2.2)axbyx<y or (x=y and ab).

Moreover, ×HM is self-similar if and only if H is self-similar.

Proof.

We show first that the multiplication (2.1) is associative. So we have to verify

(2.3)(axby)cz=ax(bycz)

for a,b,cH and x,y,zM. For x=1, this follows by the associativity of H. Assume that x1. Since xby1, the left-hand side of equation (2.3) becomes a(xby)cz=a((xby)cz). If y=1, the right-hand side is ax(bc)z=a(xbcz). So the equation follows since xbc=(xb)c. Similarly, for x,y1, equation (2.3) follows by (xby)c=(xb)(yc). Furthermore, the multiplication (2.1) admits a neutral element. Thus ×HM is a monoid.

With respect to the natural ordering (1.4), axby means that ax=czby holds for some cH and zM. Here z=1 yields x=y and ab, while z1 gives x<y. This proves (2.2), which shows, in particular, that (1.4) endows ×HM with a partial order.

In what follows, let x(ax) denote the inverse of the action xxa, that is,

axa=x=(ax)a

for aH and xM{1}. Now we define a second operation on ×HM by

(2.4)axby:={a(xy)for y⩽̸x⩽̸y,b(a(xy))for y<x,abfor y=x,1for x<y.

Next we verify the equivalence

(2.5)axbyczaxbycz

in ×HM. Assume that x=1. Then the left-hand side states that (ab)ycz, that is, y<z or (y=z and abc). For y<z, both sides of (2.5) are true, while for y=z, both sides of (2.5) are equivalent to abc. So we can assume that x1. The left-hand side of (2.5) is then equivalent to a(xby)cz, that is,

(2.6)xby<zor(xby=z and ac).

If y and z are incomparable, bycz=b(yz). So the right-hand side of (2.5) becomes xb(yz), that is, xbyz. Since z⩽̸y, this is equivalent to xby<z. For z<y, the right-hand side of (2.5) is axc(b(yz)), which gives either xb<yz or (xb=yz and ac). Since M is self-similar, xb=yz is equivalent to xby=z. So the right-hand side of (2.5) can be transformed into (2.6). Finally, since x1, both sides of (2.5) are true if yz. This completes the proof of (2.5).

By Lemma 3, ×HM is a left hoop. Using Proposition 1, it remains to verify that H is self-similar if and only if

(2.7)axbyax=by

holds in ×HM. For y=1, we have axbyax=ax(ba)x=aba. So equation (2.7) becomes aba=b. If y1, then

axbyax=axb(yax)
=b(a(x(ya)x))
=b(aya)=by.

To any totally ordered abelian group G we associate a self-similar left hoop Lex(G) as follows. Let M(G) be the free monoid over G. To distinguish the multiplication of M(G) with that of G, we write the elements of M(G) as a1.a2..an. As G is an additive group, there is no confusion between the neutral element 0 of G and the unit element of M(G), the empty word 1. We define Lex(G) to be the free monoid over G with the relation

(2.8)a.b=a+bfor a<0.
Proposition 7

Let G be a totally ordered abelian group. Then Lex(G) is a self-similar left hoop.

Proof.

Using equation (2.8), the elements of Lex(G) can be put into the normal form a1.a2..an with a1,,an-10. The existence of such a form is trivial. Uniqueness follows by considering the maximal intervals [i,k] with ai++aj<0 for all j{i,,k}. Thus, from now on, we represent the elements of Lex(G) by their normal form. Multiplying a1.a2..an from the left by some aG+, we get a.a1.a2..an. If n2, multiplying by -a1-a for some a>0 yields (a2-a).a3..an. So the natural ordering (1.4) provides Lex(G) with a linear, lexicographic order. Note, however, that the map s:Lex(G) with

s(a1.a2..an):=a1++an

is neither increasing nor decreasing.

For u,vLex(G) and aG, it is easily checked that ua=va implies u=v. Hence Lex(G) is right cancellative. It follows that, if uv, there is a unique uvLex(G) with v=(uv)u. If uv, we set uv:=1. Hence we have (uv)u=uv. In particular, uvu=v holds for all u,vLex(G). Since uvuwvw, it follows that uwvw=(uv)w, that is, we have (uwvw)uw=(uv)uw. Hence uwvw=uv. Thus

(uvw)uv=uvw
=uvvw
=(uv(vw)v)uv
=(u(vw))uv,

which yields uvw=u(vw). ∎

Example 4

Let Xa be the L-algebra of Example 1. Consider the monoid H generated by a,b with a single relation b2a=b. It is easily checked that the elements of H can be put into the normal form

u=amban1+1ban2+1banr+1bn

with r,m,n,n1,,nr. The elements with m=0 form a submonoid M. To any uM, we associate the sequence (n1,,nr) if n=0, and (n1,,nr,-n) if n>0. Then a straightforward calculation shows that M is isomorphic to Lex(), and

H×S(Xa)Lex(),

a self-similar left hoop with structure group . The canonical map q:H is given by

um+n1++nr-n.

In general, the image of the map q:HG(H) is a submonoid of G(H), but q(H)q(H)-1 need not be trivial. So the partial order of H need not induce a partial order of G(H). In the present example, q(H)=q(H)-1=G(H).


Dedicated to B. V. M.



Communicated by Christopher W. Parker


Acknowledgements

The author thanks the referee for helpful remarks which led to an improvement of the paper.

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Received: 2015-10-14
Revised: 2016-7-1
Published Online: 2016-9-20
Published in Print: 2017-3-1

© 2017 by De Gruyter

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