Garside combinatorics for Thompson’s monoid $F^+$ and a hybrid with the braid monoid $B_{\infty }^{+}$

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ALGEBRAIC COMBINATORICS

Patrick Dehornoy & Emilie Tesson

Garside combinatorics for Thompson’s monoidF⁺and a hybrid with the braid monoidB_∞⁺

Volume 2, issue 4 (2019), p. 683-709.

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Garside combinatorics for Thompson’s monoid F ⁺ and a hybrid with the braid

monoid B _∞ ⁺

Patrick Dehornoy & Emilie Tesson

Abstract On the model of simple braids, defined to be the left divisors of Garside’s elements ∆nin the monoidB∞⁺, we investigate simple elements in Thompson’s monoidF⁺and in a larger monoidH⁺ that is a hybrid ofB∞⁺ and F⁺: in both cases, we count how many simple elements left divide the right lcm of the firstn−1 atoms, and characterize their normal forms in terms of forbidden factors. In the case ofH⁺, a generalized Pascal triangle appears.

1. Introduction

Since the seminal work of F.A. Garside [18], as extended in [16] and [4], it is known that Artin’s braid group Bn is a group of fractions for the monoid B_n⁺ of positive n-strand braids and that the now called Garside element ∆n plays a prominent role in the study of B_n⁺. In particular, the divisors of ∆n in B_n⁺, called simple braids, form a family ofn! elements in one-to-one correspondence with the permutations of {1, . . . , n}, leading to a remarkable combinatorics now at the heart of the algebraic study of Bn [1, 17], see [15, Chapter IX]. Subsequently, it was realised that such a situation can be found in many different contexts of groups and categories, always around a family of so-called simple elements resembling simple braids, and leading to various combinatorics, like, for instance, the dual Garside structure onB_n [3], whose combinatorics is that of noncrossing partitions.

Our aim in this paper is to investigate a Garside structure arising on Thompson’s group F [27, 8] in connection with its submonoid F⁺ generated by the standard (infinite) sequence of generators, corresponding to the presentation

(1) F⁺:=hτ1, τ2, . . .|τjτi=τiτj+1 for j>i+ 1i⁺.

To explain the similarity with braids and the natural questions in this non-finitely generated case, one should start from the infinite braid monoid

(2) B⁺_∞=

σ₁, σ₂, . . .

σ_jσ_i=σ_iσ_j forj>i+ 2 σ_jσ_iσ_j =σ_iσ_jσ_i forj=i+ 1

⁺ :

in this case, Garside’s braid ∆n is the right lcm of then−1 first atomsσ₁, . . . , σ_n−1 ofB_∞⁺ (see Section 2.1 for a reminder about the terminology), and simple braids are those braids that left divide at least one element ∆n in B_∞⁺.

Manuscript received 8th March 2018, accepted 28th November 2018.

Keywords. presented monoid, divisibility relation, simple elements, Thompson’s group, braid group, normal form, Garside element, directed animal.

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In the case of the monoid F⁺, the atoms are the elements τ_i, and we shall see that there exists for each n a well defined element ∆_n that is, in F⁺, the right lcm of the first n−1 atoms. Then we shall investigate the derived simple elements, namely the elements ofF⁺that left divide at least one element ∆n. The main results proved here are that, for everyn, there exist 2ⁿ⁻¹ simple elements left dividing ∆n

in F⁺, in explicit one-to-one correspondence with the subsets of {1, . . . , n−1}, and that simple elements form a Garside family in F⁺ (the definition is recalled below, before Corollary 2.15), thus guaranteeing the existence and properties of an associated greedy normal form inF⁺. These results are established by combining the existence of a convergent rewrite system onF⁺ and the reversing technique [10, 12] for analyzing the divisibility relations of a presented monoid.

The above results are technically easy, and we then switch to a combinatorially more involved situation related to another monoidH⁺, which is a hybrid of the braid monoidB⁺_∞and the Thompson monoidF⁺. Various hybrids of the groupsB_∞andF have already been considered, in particular the groupBVdof [7, 6, 11], which is a group of fractions for a monoid, that is a Zappa–Szép product of F⁺ and B_∞⁺ [5, 24, 29]

and, therefore, possibly inherits their Garside structures. Here we shall introduce and investigate a new hydrid, which is not a product but rather a mixture of the initial monoidsF⁺ andB_∞⁺. Indeed, we consider

(3) H⁺:=

θ1, θ2, . . .

θjθi=θiθj+1 forj>i+ 2 θjθiθj =θiθjθi+3 forj=i+ 1

+

,

in which the length 2 relations are Thompson’s relations as in (1), whereas the length 3 relations are directly reminiscent of braid relations of (2), but with a shift of one index; the “mysterious” value i+ 3 is chosen to guarantee that F⁺ is a quotient ofH⁺. Here, we investigate the basic properties of the monoidH⁺ and, specifically, the associated Garside combinatorics, if this makes sense. Actually, it does: we shall see that, for everyn, the atomsθ₁, . . . , θ_n−1admit a right lcm, again denoted by ∆n, so that it is natural to investigate simple elements, defined to be those that left divide some element ∆n. The main results proved here are that, for every n, there exist 2·3ⁿ⁻² simple elements left dividing ∆n in H⁺, with an explicit description of a distinguished expression for each of them. As in the case of F⁺, these results are established using a convergent rewrite system on H⁺ and the reversing technique;

the proofs are more difficult than forF⁺ and some of them require delicate inductive arguments. We hope that the existence of this nontrivial combinatorics will draw some attention to the monoid H⁺, and to the group H presented by (3), which remains essentially mysterious.

The paper is divided into four sections after this introduction. In Section 2, we investigate the monoidF⁺and the derived simple elements, providing a good warm-up for the sequel. In Section 3, we establish various general properties of the monoidH⁺, in particular the fact that it admits cancellation on both sides. Next, in Section 4, we study the elements ∆_n ofH⁺ and count their left divisors by partitioning them into several families. Finally, in Section 5, we explicitly characterize the normal form (in the sense of some convergent rewrite system) of simple elements ofH⁺.

2. Thompson’s monoid F⁺

Here we study the case of Thompson’s monoid F⁺, an easy first step. It is standard that (1) is a presentation of Thompson’s group F, and, as the relations involve no inverse of the generators, it makes sense to introduce the associated monoidF⁺ and to consider the associated Garside combinatorics, if it exists.

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The section is divided into four parts. In Section 2.1, we recall the standard terminology for the divisibility relations in a monoid, extensively used throughout the text. Next, in Section 2.2, we define a convergent rewrite system that selects a distinguished expression for every element ofF⁺. In Section 2.3, we recall basic notions about word reversing, here in the new version of [14], and use them to show that F⁺ is cancellative and admits right lcms (least common right multiples). Finally, in Section 2.4, we investigate the elements ∆n and describe their left divisors explicitly.

2.1. The divisibility relations of a monoid. LetM be a monoid (possibly, in particular, a free one, i.e. a monoid of words). Fora, binM, we say thataleft dividesb in M, or, equivalently, that b is aright multiple ofa, written a4b, if ax=b holds for somex(ofM). IfM is left cancellative (meaning thatxa=xbimpliesa=b) and 1 is the only invertible element inM, the relation4is a partial ordering on M.

Fora, binM, we say thatcis aright lcm(least common right multiple) ofaandb ifa 4c and b 4c hold, and the conjunction of a4x and b 4ximplies c 4x: in other words,cis a lowest upper bound of aandbwith respect to 4.

The symmetric notions of a right divisor and a left multiple are defined similarly, replacingax=b withxa=b. Finally, we say thatais afactor ofbifxay=b holds for somex, y.

An elementaofM is said to be anatom if it admits no decompositiona=bcwith b6= 1 andc6= 1.

2.2. A normal form on F⁺. We begin our investigation of the monoid F⁺. We recall thatF⁺ is defined by the presentation

F⁺:=hτ1, τ2, . . .|τjτi=τiτj+1 for j>i+ 1i⁺,

hereafter denoted by PF. We put T := {τi | i > 1}, write T^∗ for the free monoid of all words in the alphabet T, and ≡ for the congruence on T^∗ generated by the relations ofPF. We useεfor the empty word. Our first tool for studyingF⁺ consists in defining a unique normal form using a rewrite system onT^∗.

Lemma 2.1.Let EF be the rewrite system on T^∗ defined by the rules (4) τiτj+1→τjτi fori>1 andj>i+ 1.

ThenE_F is convergent.

Proof. As is standard, see for instance [25], we shall check thatEF is noetherian and locally confluent. We write ⇒ for the one-step rewrite relation associated with the rules of (4), that is, for the family of all pairs

(w₁τ_iτ_j+1w₂ , w₁τ_jτ_iw₂) withj>i+ 1,

and⇒^∗ for the reflexive-transitive closure of⇒. ForwinT^∗, letρ(w) be the sum of the indices of the generatorsτi occurring in w. Thenw⇒w⁰ impliesρ(w)> ρ(w⁰), and, therefore, there is no proper infinite sequence for⇒. SoEF is noetherian.

Next, assumew⇒w⁰ andw⇒w⁰⁰. By definition,w⁰ andw⁰⁰ are obtained fromw by replacing some length 2 factorτiτj+1 with the corresponding word τjτi. For local confluence, the case of disjoint factors is trivial, and the critical case of overlapping factors corresponds to w = τiτj+1τk+2 with j > i+ 1 and k > j + 1, leading to w⁰=τjτiτk+2and w⁰⁰=τiτk+1τj+1. One then obtains

(5) τiτj+1τk+2

τ_jτ_iτ_k+2

τ_iτ_k+1τ_j+1

τkτjτi ,

⇒

⇒²

⇒₂

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It follows that E_F is locally confluent, hence convergent by Newman’s diamond

lemma [21].

For every wordwofT^∗, we shall denote byred(w) the uniqueEF-reduced wordw⁰ satisfyingw⇒^∗w⁰. By definition, the wordswandred(w) represent the same element ofF⁺, andred(w) is the uniqueEF-reduced word in the equivalence class ofwinF⁺. Thus, Lemma 2.1 implies

Proposition 2.2.EF-reduced words provide a unique normal form for the elements of the monoidF⁺.

It directly follows from the definition that a word ofT^∗ isEF-reduced if, and only if, it has no length 2 factorτiτj+1 withj>i+ 1, which implies that, for everyn, the set ofEF-reduced words lying in{τ1, . . . , τn}^∗ is a regular language [17, 20].

2.3. Using word reversing. The second method for investigating the monoidF⁺ is word reversing [12], a distillation of an argument that ultimately stems from Gar- side’s approach to braid monoids [18]. Here we shall describe reversing using the new formalism of [14], which is specially convenient in the current case (and in that ofH⁺in Section 3.3). So we introduce reversing as a binary relation on pairs of words connected with a particular type of van Kampen diagram.

Definition 2.3.[14] A reversing grid for a monoid presentation (S,R), or (S,R)- grid, is a rectangular diagram consisting of finitely many matching S ∪ {ε}-labeled pieces of the types

-

t s

t1 tq

s1

sp

withs, t, s1, ..., sp, t1, ..., tq inS

andst1···tq=ts1···spa relation of R,

- s s

ε ε,

ε s

ε s,

t ε

t ε,

ε ε

ε

ε withs, tinS.

Foru, v, u₁, v₁ in S^∗, we say that an(S,R)-grid Γ goesfrom (u, v) to (u₁, v₁)if the labels of the left and top edges ofΓform the wordsuandv, respectively, whereas the labels of the right and bottom edges form the words u₁ andv₁. We write (u, v) y^R (u₁, v₁)if there exists a(S,R)-grid from(u, v)to(u₁, v₁).

Example 2.4.Two typicalPF-grids are

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τ1 τ3

τ1 ε

τ2 τ3 ε,

τ2 τ1

ε τ1

τ2 ε ε,

witnessing for the relations (τ2, τ1τ3) y(ε, τ1) and (τ2, τ2τ1)y (ε, τ1), respectively (we omit the index inywhen there is no ambiguity). Note that, because all relations of PF involve words of length 2, the pieces of the first type in Definition 2.3 are squares: the right and bottom edges each consist of one singleS-labeled arrow.

The following result is (a special case of a result) established in [14]. Below we say that a monoid presentation (S,R) ishomogeneous if every relation inRhas the formw=w⁰ withw, w⁰ of the same length, andright complemented if it contains no relations . . .=s . . .and at most one relation s . . .=t . . . for alls6=t in S. On the other hand, two (S,R)-grids Γ from (u, v) to (u1, v1) and Γ⁰ from (u⁰, v⁰) to (u⁰₁, v₁⁰)

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Garside combinatorics forF⁺ and a hybrid withB⁺∞

are equivalent if we have u⁰ ≡_R u, v⁰ ≡_R v, u⁰₁ ≡_R u₁, andv⁰₁ ≡_R v₁, where≡_R is the congruence onS^∗ generated byR—so that the monoidhS | Ri⁺ isS^∗/≡_R. Lemma2.5 ([14, Propositions 1.12, 1.14, 1.16]).Assume that(S,R)is a homogeneous right complemented monoid presentation and, for every s in S and every relation w=w⁰ inR,

(♦) for every grid from (s, w), there is an equivalent grid from(s, w⁰),

and vice versa.

(i) Two wordsu, v of S^∗ represent the same element of the monoidhS | Ri⁺ if, and only if,(u, v)y(ε, ε)holds.

(ii) The monoidhS | Ri⁺ is left cancellative.

(iii) Two elementsa, bof hS | Ri⁺ represented byuandv inS^∗ admit a common right multiple if, and only if, (u, v) y^R (u1, v1) holds for some u1, v1; in this case, the element represented by uv1 is a right lcm of a and b. In the special case when, for alls6=t inS, there exists⁰, t⁰ inS such that st⁰=ts⁰ is a relation of R, there always existu₁, v₁ as above, and any two elements ofhS | Ri⁺ admit a right lcm.

Applying Lemma 2.5, we deduce:

Proposition 2.6.The monoid F⁺ is left and right cancellative. Any two elements ofF⁺ admit a right lcm. Any two elements ofF⁺ that admit a common left multiple admit a left lcm.

Proof. In view of applying Lemma 2.5, we observe that the presentation P_F is homogeneous (all relations are of the formw=w⁰ withwand w⁰ of length two), right complemented with one relationτi. . .=τj. . . for alli, j, and we have to prove that Condition (♦) holds for every τi and every relation of PF. To do that, we consider all pairs (τi, τjτk+1) withk>j+ 1, and compare the reversing grids from (τi, τjτk+1) and from (τi, τkτj): the two grids of Example 2.4 are typical, corresponding toi= 2, j= 1, and k= 2, and they are indeed equivalent, since both admit as output (ε, τ1).

The number of triples (i, j, k) to consider is infinite but only finitely many patterns may occur, according to the position ofiwith respect tojandk. We skip the details, which are fairly obvious. Having established (♦), we deduce from Lemma 2.5(ii) that the monoid F⁺ is left cancellative and from Lemma 2.5(iii) that any two elements ofF⁺ admit a right lcm.

To study left multiples, we observe that the presentation PF is also left complemented (in the obvious sense), and consider the notion of a left reversing grid, which is symmetric to the above notion of a right reversing grid (which amounts to considering the opposed monoid). To this end, we replace each elementary dia-

gram

The following result is (a special case of a result) established in [9]. Below we say that a monoid presentation (S,R) is homogeneous if every relation inRhas the form w =w^′ with w, w^′ of the same length, and right complemented if it contains no relations... =s... and at most one relations... =t... for alls 6=t in S. On the other hand, two (S,R)-grids Γ from (u, v) to (u1, v1) and Γ^′ from (u^′, v^′) to (u^′₁, v₁^′) areequivalent if we haveu^′ ≡R u, v^′ ≡R v, u^′₁ ≡R u1, andv₁^′ ≡R v1, where ≡R is the congruence onS^∗generated byR—so that the monoidhS | Ri⁺ isS^∗/≡R. Lemma 2.5. [9, Propositions 1.12, 1.14, 1.16] Assume that (S,R) is a homogeneous right complemented monoid presentation and, for every s in S and every relation w=w^′ in R,

(♦) for every grid from (s, w), there is an equivalent grid from(s, w^′), and vice versa.

(i)Two wordsu, vofS^∗ represent the same element of the monoidhS | Ri⁺if, and only if,(u, v)y(ε, ε)holds.

(ii)The monoid hS | Ri⁺ is left cancellative.

(iii)Two elements a, b ofhS | Ri⁺ represented by u andv in S^∗ admit a common right multiple if, and only if, (u, v) yR (u1, v1) holds for some u1, v1; in this case, the element represented byuv1is a right lcm ofaandb. In the special case when, for alls6=t inS, there exists^′, t^′ inS such thatst^′=ts^′ is a relation ofR, there always exist u1, v1 as above, and any two elements of hS | Ri⁺ admit a right lcm.

Proposition 2.6. The monoid F⁺ is left and right cancellative. Any two elements ofF⁺ admit a right lcm. Any two elements ofF⁺ that admit a common left multiple admit a left lcm.

Proof. In view of applying Lemma 2.5, we observe that the presentationP^F is homogeneous (all relations are of the formw=w^′ with wandw^′ of length two), right complemented with one relation τi... = τj... for all i, j, and we have to prove that Condition (♦) holds for everyτi and every relation ofP^F. To do that, we consider all pairs (τi, τjτk+1) withk>j+ 1, and compare the reversing grids from (τi, τjτk+1) and from (τi, τkτj): the two grids of Example 2.4 are typical, corresponding toi= 2, j= 1, andk= 2, and they are indeed equivalent, since both admit as output (ε, τ1).

The number of triples (i, j, k) to consider is infinite but only finitely many patterns may occur, according to the position ofiwith respect tojandk. We skip the details, which are fairly obvious. Having established (♦), we deduce from Lemma 2.5(ii) that the monoid F⁺ is left cancellative and from Lemma 2.5(iii) that any two elements ofF⁺ admit a right lcm.

To study left multiples, we observe that the presentation P^F is also left complemented (in the obvious sense), and consider the notion of a left reversing grid, which is symmetric to the above notion of a right reversing grid (which amounts to considering the opposed monoid). To this end, we replace each elementary dia-

gram

t s

t1 tq

s1

sp

of Definition 2.3 with its counterpart

t

s t1 tq

s1

sp

fors1···spt =t1···tqs in Rand, similarly, replace s s

ε

ε with ε ε

s

s. Then one easily checks that the counterpart of (♦) is satisfied and one deduces, by the

Algebraic Combinatorics, draft (February 28, 2019) 5

The following result is (a special case of a result) established in [9]. Below we say that a monoid presentation (S,R) ishomogeneous if every relation inRhas the form w = w^′ with w, w^′ of the same length, and right complemented if it contains no relations... =s... and at most one relations... = t...for alls 6=t in S. On the other hand, two (S,R)-grids Γ from (u, v) to (u1, v1) and Γ^′ from (u^′, v^′) to (u^′₁, v₁^′) areequivalent if we haveu^′ ≡R u, v^′ ≡R v, u^′₁ ≡R u1, andv^′₁ ≡R v1, where≡R is the congruence onS^∗ generated byR—so that the monoidhS | Ri⁺ isS^∗/≡R. Lemma 2.5. [9, Propositions 1.12, 1.14, 1.16] Assume that (S,R) is a homogeneous right complemented monoid presentation and, for every s in S and every relation w=w^′ inR,

(♦) for every grid from(s, w), there is an equivalent grid from (s, w^′), and vice versa.

(i)Two wordsu, vofS^∗ represent the same element of the monoidhS | Ri⁺ if, and only if,(u, v)y(ε, ε)holds.

(ii)The monoid hS | Ri⁺ is left cancellative.

(iii)Two elements a, b ofhS | Ri⁺ represented by u andv in S^∗ admit a common right multiple if, and only if, (u, v) yR (u1, v1) holds for some u1, v1; in this case, the element represented byuv1is a right lcm ofaandb. In the special case when, for alls6=tinS, there exists^′, t^′ inS such thatst^′=ts^′ is a relation of R, there always exist u1, v1 as above, and any two elements of hS | Ri⁺ admit a right lcm.

Proposition 2.6. The monoid F⁺ is left and right cancellative. Any two elements ofF⁺ admit a right lcm. Any two elements ofF⁺ that admit a common left multiple admit a left lcm.

Proof. In view of applying Lemma 2.5, we observe that the presentationP^F is homogeneous (all relations are of the formw=w^′ withwandw^′ of length two), right complemented with one relation τi... = τj... for all i, j, and we have to prove that Condition (♦) holds for everyτi and every relation ofP^F. To do that, we consider all pairs (τi, τjτk+1) withk>j+ 1, and compare the reversing grids from (τi, τjτk+1) and from (τi, τkτj): the two grids of Example 2.4 are typical, corresponding toi= 2, j= 1, andk= 2, and they are indeed equivalent, since both admit as output (ε, τ1).

The number of triples (i, j, k) to consider is infinite but only finitely many patterns may occur, according to the position ofiwith respect tojandk. We skip the details, which are fairly obvious. Having established (♦), we deduce from Lemma 2.5(ii) that the monoid F⁺ is left cancellative and from Lemma 2.5(iii) that any two elements ofF⁺admit a right lcm.

gram

t s

t1 tq

s1

sp

t

s t1 tq

s1

sp

fors1···spt =t1···tqs in Rand, similarly, replace s ε

ε with s

one easily checks that the counterpart of (♦) is satisfied and one deduces, by the

for

s₁· · ·s_pt =t₁· · ·t_qs in R and, similarly, replace

The following result is (a special case of a result) established in [9]. Below we say that a monoid presentation (S,R) ishomogeneous if every relation inRhas the form w = w^′ with w, w^′ of the same length, and right complemented if it contains no relation s... =s... and at most one relation s... =t... for alls 6=t in S. On the other hand, two (S,R)-grids Γ from (u, v) to (u1, v1) and Γ^′ from (u^′, v^′) to (u^′₁, v^′₁) are equivalent if we haveu^′ ≡R u, v^′ ≡R v, u^′₁ ≡R u1, and v₁^′ ≡R v1, where ≡R is the congruence onS^∗ generated byR—so that the monoidhS | Ri⁺ isS^∗/≡R. Lemma 2.5. [9, Propositions 1.12, 1.14, 1.16] Assume that (S,R) is a homogeneous right complemented monoid presentation and, for every s in S and every relation w=w^′ inR,

(i)Two wordsu, vofS^∗represent the same element of the monoidhS | Ri⁺ if, and only if, (u, v)y(ε, ε)holds.

(ii) The monoid hS | Ri⁺ is left cancellative.

(iii) Two elements a, bof hS | Ri⁺ represented by uandv inS^∗ admit a common right multiple if, and only if, (u, v)yR (u1, v1) holds for some u1, v1; in this case, the element represented byuv1 is a right lcm ofaandb. In the special case when, for alls6=tinS, there exists^′, t^′ inS such thatst^′ =ts^′ is a relation ofR, there always exist u1, v1 as above, and any two elements of hS | Ri⁺ admit a right lcm.

Proposition 2.6. The monoid F⁺ is left and right cancellative. Any two elements of F⁺ admit a right lcm. Any two elements of F⁺that admit a common left multiple admit a left lcm.

Proof. In view of applying Lemma 2.5, we observe that the presentation P^F is homogeneous (all relations are of the form w=w^′ withw andw^′ of length two), right complemented with one relation τi... = τj... for all i, j, and we have to prove that Condition (♦) holds for every τi and every relation of P^F. To do that, we consider all pairs (τi, τjτk+1) withk>j+ 1, and compare the reversing grids from (τi, τjτk+1) and from (τi, τkτj): the two grids of Example 2.4 are typical, corresponding to i= 2, j = 1, andk= 2, and they are indeed equivalent, since both admit as output (ε, τ1).

The number of triples (i, j, k) to consider is infinite but only finitely many patterns may occur, according to the position ofiwith respect toj andk. We skip the details, which are fairly obvious. Having established (♦), we deduce from Lemma 2.5(ii) that the monoid F⁺ is left cancellative and from Lemma 2.5(iii) that any two elements ofF⁺ admit a right lcm.

gram

t

s s1

sp

of Definition 2.3 with its counterpart s t1 tq

s1

sp

s s

ε ε ith

ε ε

s

with

The following result is (a special case of a result) established in [9]. Below we say that a monoid presentation (S,R) ishomogeneous if every relation inRhas the form w = w^′ with w, w^′ of the same length, and right complemented if it contains no relations... =s... and at most one relation s... =t... for alls 6=t in S. On the other hand, two (S,R)-grids Γ from (u, v) to (u1, v1) and Γ^′ from (u^′, v^′) to (u^′₁, v^′₁) areequivalent if we have u^′ ≡R u, v^′ ≡R v, u^′₁ ≡R u1, andv₁^′ ≡R v1, where ≡R is the congruence onS^∗ generated byR—so that the monoidhS | Ri⁺ isS^∗/≡R. Lemma 2.5. [9, Propositions 1.12, 1.14, 1.16] Assume that (S,R) is a homogeneous right complemented monoid presentation and, for every s in S and every relation w=w^′ inR,

(i)Two wordsu, vofS^∗represent the same element of the monoidhS | Ri⁺ if, and only if,(u, v)y(ε, ε)holds.

(ii) The monoid hS | Ri⁺ is left cancellative.

(iii) Two elements a, bof hS | Ri⁺ represented by uandv inS^∗ admit a common right multiple if, and only if, (u, v) yR (u1, v1) holds for some u1, v1; in this case, the element represented byuv1 is a right lcm ofaandb. In the special case when, for alls6=t inS, there exists^′, t^′ inS such thatst^′ =ts^′ is a relation ofR, there always existu1, v1 as above, and any two elements ofhS | Ri⁺ admit a right lcm.

Proposition2.6. The monoid F⁺ is left and right cancellative. Any two elements ofF⁺admit a right lcm. Any two elements of F⁺ that admit a common left multiple admit a left lcm.

Proof. In view of applying Lemma 2.5, we observe that the presentation P^F is homogeneous (all relations are of the formw=w^′ withw andw^′ of length two), right complemented with one relationτi... =τj... for all i, j, and we have to prove that Condition (♦) holds for every τi and every relation of P^F. To do that, we consider all pairs (τi, τjτk+1) withk>j+ 1, and compare the reversing grids from (τi, τjτk+1) and from (τi, τkτj): the two grids of Example 2.4 are typical, corresponding toi= 2, j= 1, andk= 2, and they are indeed equivalent, since both admit as output (ε, τ1).

The number of triples (i, j, k) to consider is infinite but only finitely many patterns may occur, according to the position ofiwith respect toj andk. We skip the details, which are fairly obvious. Having established (♦), we deduce from Lemma 2.5(ii) that the monoid F⁺ is left cancellative and from Lemma 2.5(iii) that any two elements ofF⁺ admit a right lcm.

gram

t s

t1 tq

s1

sp

of Definition 2.3 with its counterpart s t1 tq

s1

sp

fors1···spt =t1···tqsin Rand, similarly, replace s s

ε

ε ε

s

. Then one easily checks that the counterpart of (♦) is satisfied and one deduces, by the counterpart of Lemma 2.5(ii), that F⁺ is right cancellative. Finally, the counterpart of Lemma 2.5(iii) implies that any two elements ofF⁺ that admit a common left multiple admit a left lcm. However, two elements ofF⁺ need not always admit a common

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left multiple: there is no relation. . . τ₁=. . . τ₂inP_F, and, therefore, the counterpart of Lemma 2.5(iii) implies thatτ₁ andτ₂admit no common left-multiple inF⁺. It follows from Proposition 2.6 and Ore’s classical theorem [22] that the monoidF⁺ embeds in its enveloping group, which is the group presented byPF, namely Thomp- son’s groupF, and that the latter is a group of right fractions forF⁺, that is, every element ofF can be expressed asab⁻¹ with a, bin F⁺. The expression is unique if, in addition, we require that the fraction be irreducible, meaning thataand b admit no common right divisor.

Remark2.7.As explained in [13], there exists a (more redundant) positive presenta- tionP_F^∗ of the groupF in terms of a family of generatorsτ_s^∗ withsa finite sequence of 0s and 1s such thatτi coincides withτ₁^∗i−1 and thatF is a group both of left and right fractions for the monoidF^+∗ defined by P_F^∗. The latter admits left and right lcms and is a sort of counterpart for the dual braid monoid of [3]. The main relations inP_F^∗ correspond to the MacLane–Stasheff pentagon.

2.4. Garside combinatorics for F⁺. The monoidF⁺ resembles the braid mon- oidB_∞⁺ in that it is cancellative and admits right lcms and, therefore, it makes sense to consider the counterpart of the Garside elements ∆n and their divisors.

As the presentationPF is homogeneous, the atoms ofF⁺ are the elementsτi with i > 1. So, exactly as in the case of B_∞⁺, we shall consider the element ∆n that is the right lcm ofτ1, . . . , τ_n−1—we might use a different notation, for instance ∆^F_n, but there will be no risk of ambiguity here. We start from an explicit expression.

Definition 2.8.We put∆₁:=ε, and, for n>2, we put∆_n :=τ1τ3τ5· · ·τ_2n−3. We denote by∆n the class of∆_n in F⁺.

It is clear that ∆n left divides ∆n+1 for each n, and one inductively checks that theEF-normal form of ∆n isτ_n−1τ_n−2· · ·τ2τ1.

Lemma 2.9.For every n > 2, the element ∆n is the right lcm of τ1, . . . , τ_n−1. No elementτi with i>nleft divides∆n.

Proof. We prove using induction on n>2 that ∆_n is the right lcm of τ₁, . . . , τ_n−1. The result is trivial forn= 2. Assumen>3. A direct computation gives

(τ_n−1,∆_n−1)y(τ_2n−3,∆_n−1).

By Lemma 2.5(iii), this implies that ∆_n represents the right lcm ofτn−1 and ∆n−1. By induction hypothesis, ∆n−1is the right lcm ofτ₁, . . . , τ_n−2, so ∆n is the right lcm ofτ₁, . . . , τ_n−1. On the other hand, for i>n, we find (τ_i,∆_n)y(τ_i+n−1,∆_n), which shows that the right lcm ofτ_i and ∆_n is not ∆_n, soτ_i does not left divide ∆_n. The main notion in Garside theory [15] is the notion of a simple element, defined as the (left) divisors of the distinguished element(s) ∆.

Definition 2.10.An element aofF⁺ is called simpleif a4∆_n holds for some n.

Our aim is to understand the structure of simple elements of F⁺, typically to characterize their normal forms. To this end, the key point will be the following exhaustive description of the expressions of ∆_n. Below, we write S_n for the group of all permutations of {1, . . . , n}, and s_i for the transposition (i, i+ 1). The result should be seen as a counterpart of the (obvious) result that, in a free commutative monoid with atomsa₁, . . . , a_n−1, the expressions of the product ∆_n:=a₁· · ·a_n−1are obtained by permuting the letters, hence in one-to-one correspondence withS_n−1. In the case ofF⁺, shifting the switched indices requires to use the dilated versionfeof a permutationf introduced in (7).

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Lemma2.11.The expressions of∆_n are the wordsw_f withf inS_n−1, where, forp6 n−1, we put

(7) fb(p) := #{i < f⁻¹(p)|f(i)> p} and fe(p) := 2f⁻¹(p)−1−fb(p), and letw_f be the wordτ

f(1)e τ

f(2)e · · ·τ

f(n−1)e .

Proof. We first establish the following technical result:

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Iff⁻¹(p)< f⁻¹(p+ 1) (resp.>) holds, then so doesfe(p) + 1<fe(p+ 1) (resp.fe(p) >fe(p+ 1)); applying a relation of PF to wf in position p yields the wordws_pf.

In order to prove (8), assumef⁻¹(p+ 1) = f⁻¹(p) +mwith m>1. The definition gives

fb(p+ 1) =f(p) + #{ib |f⁻¹(p) p+ 1},

whence fb(p+ 1)6fb(p) +m−1 and, for there,fe(p+ 1) >fe(p) + 2. Let g :=s_pf. We find g⁻¹(p) = f⁻¹(p+ 1), g⁻¹(p+ 1) = f⁻¹(p), then g(p) =b fb(p+ 1) + 1 and bg(p+ 1) =fb(p), becausef⁻¹(p) contributes tobg(p) but not tofb(p+ 1), and, finally, eg(p) =fe(p+ 1)−1 andeg(p+ 1) =fe(p), with eg(q) =fe(q) forq6=p, p+ 1. So wg is the result of applying the ruleτ

f(p)e τ

f(p+1)e →τ

f(p+1)−1e τ

f(p)e towf in positionp.

On the other hand, for f⁻¹(p) = f⁻¹(p+ 1) +m with m > 1, we find fb(p) 6 fb(p+ 1) +m, leading tofe(p)>fe(p+ 1) + 1. Forg:=spf, we find now,bg(p) =fb(p+ 1) andg(p+1) =b fb(p)−1, whenceeg(p) =fe(p+1) andeg(p+1) =fe(p)+1, witheg(q) =fe(q) forq6=p, p+ 1. Sow_g is the result of applying the ruleτ

f(p)e τ

f(p+1)e →τ

f(p+1)e τ

f(p)+1e tow_f in positionp.

Now, (8) implies that the family W := {wf | f ∈ S_n−1} is closed under ≡. As the transpositionss_i generateS_n−1, this familyW is the≡-equivalence class of the

wordw_id, which, by definition, is ∆_n.

From there, a complete description of simple elements ofF⁺ follows:

Proposition 2.12.For everyain F⁺, the following are equivalent:

(i) The elementais simple, i.e.aleft divides some element ∆n; (ii) The elementais a factor of some element ∆n;

(iii) The normal form ofa has the formτi1· · ·τi_` with i1>· · ·> i`.

Moreover,aleft divides ∆n if, and only if,nf(a)isτi₁· · ·τi_` with n > i1>· · ·> i`. Proof. By definition, (i) implies (ii). Next, assume that a is a factor of ∆n, say

∆n =a1aa2. Let w1, w, w2 be the normal forms ofa1,a, and a2, respectively. Then w1ww2 is an expression of ∆n, so, by Lemma 2.11(ii), it is a wordwf for some permutationf. Moreover, becausewisEF-reduced, no rule ofEF may apply to it: by (8), this implies that the indices of the generatorsτ_i inwmake a decreasing sequence. So (ii) implies (iii).

Assume now w =τ_i₁· · ·τ_i_` with i₁ > · · · > i_`. By inserting intermediate letters when i_p >i_p+1+ 2, we obtain a word w⁰ that is the normal form of ∆_i₁₊₁. Then, repeatedly applying to w⁰ some relations τ_iτ_j → τ_jτ_i+1, we push the new letters to the right starting with the last one and finishing with the first one. In this way, one obtains a new expression of ∆i₁+1 that begins with w. Sowis the normal form of a prefix of ∆i₁+1, hence of a simple element. Si (iii) implies (i).

For the last sentence, ifaleft divides ∆n, then so does the first generator ofnf(a):

by Lemma 2.9, the latter cannot be τi with i > n. Conversely, the above proof of (iii)⇒(i) shows thatτi₁· · ·τi_` left divides ∆i₁+1, hence ∆n forn > i1.

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Remark 2.13.As pointed out by a referee, a direct proof of Proposition 2.12 not relying on Lemma 2.11 is possible. Indeed, assume w=τ_i₁· · ·τ_i_` withi₁ >· · ·> i_`, and letw⁰ be obtained by erasing one letterτi_k. Then reducingw⁰τi_k+`−k givesw, so the element represented byw⁰ is a left divisor of the element represented byw. Thus, starting from the word ∆_n and erasing letters gives normal expressions of simple elements. Conversely, if the indices in an EF-reduced wordw are not decreasing, w must contain a factorτiτi or τiτi+1, and reducing any word of the formwv provides a word with the same factor. Hencewcannot represent a left divisor of ∆n.

Corollary 2.14.For every n, the number of left divisors of∆n inF⁺ is2ⁿ⁻¹. Proof. By the last statement in Proposition 2.12, mapping a subset of{1, . . . , n−1}to the decreasing enumeration of the corresponding elementsτ_i establishes a one-to-one correspondence betweenP({1, . . . , n−1}) and the left divisors of ∆_n. Another direct consequence is an explicit Garside family inF⁺. We recall from [18, Def. III.1.31] that a subfamily S of a (cancellative) monoid M is aGarside family if every element a of M admits an S-greedy decomposition, meaning a decomposi- tion a = a1· · ·ap with a1, . . . , ap in S, ap 6= 1, and, for each i, the entry ai is a (unique) greatest left divisor ofai· · ·aplying inS. Such a (necessarily unique) greedy decomposition can then be used in a number of applications involving the monoid and, possibly, its enveloping group. Therefore, describing explicit Garside families can be considered useful.

Corollary 2.15.Simple elements ofF⁺ make a Garside family in F⁺.

Proof. By definition, the family of simple elements inF⁺ is closed under right lcm:

the conjunction of a 4 ∆_n and b 4 ∆_p implies that the right lcm of a and b left divides ∆_max(n,p). On the other hand, a right divisor of a simple element must be a factor of some ∆n, hence, by Proposition 2.12, it is simple. By [15, Coro. IV.2.29], this implies that simple elements form a Garside family.

In the current case, the greedy decomposition associated with simple elements is directly connected with theEF-normal form: nf(a1), . . . ,nf(ap) are the maximal decreasing factors of nf(a). For instance, for nf(a) = τ4τ3τ2τ3τ1τ1τ2, the greedy decomposition has four entries, namelyτ4τ3τ2,τ3τ1, τ1, andτ2.

From there, all results involving greedy decompositions are valid inF⁺. However, this Garside structure ofF⁺ is mostly trivial, exactly parallel to the case of the free commutative monoid Z^(∞)_>0 , where simple elements also correspond to finite subsets of generators. In fact, the relations of PF are in essence a shifted version of the commutation rules of a free commutative monoid.

3. The monoid H⁺

The previous results are elementary and easy, and we now switch to a combinatorially more intricate and interesting situation, connected with the new hybridH⁺ between Thompson’s monoidF⁺and Artin’s braid monoidB_∞⁺ mentioned in the introduction.

Our aim will be to develop the same analysis as in the case of F⁺, namely under- standing the structure of simple elements, defined as the left divisors of the right lcms of atoms. To this end, we shall follow the same scheme as in Section 2 and use both a normal form associated with a rewrite system (Section 3.2) and the reversing transformation associated with the presentation (Section 3.3).

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3.1. Presentation and first properties. We recall that H⁺ is the monoid defined by the explicit presentation called (3) in the introduction

H⁺:=

θ1, θ2, . . .

θjθi=θiθj+1 forj>i+ 2 θjθiθj =θiθjθi+3 forj=i+ 1

⁺ ,

hereafter denoted byPH. We put Θ :={θi|i>1}, and write≡for the congruence on Θ^∗ generated by the relations ofPH. Forwa word of Θ^∗, we write [w] for the≡- class ofw. The relations ofPHshould appear as a mixture of the Thompson relations (as for length 2 relations), and of braid relations (as for length 3 relations). We immediately see that PH is a homogeneous presentation, and we can refer without ambiguity to the length|a|of an elementaofH⁺, defined to be the common length of all words of Θ^∗ that represent a. We also observe that the relations are invariant under shifting the indices of theθ_is by +1, implying that mappingθ_itoθ_i+1for eachi induces a well defined endomorphism ofH⁺.

Unlike the case ofB_∞⁺, the family of generators occurring in a word is not invariant under≡: for instance,θ₃θ₁is equal toθ₁θ₄. However, we can easily construct an upper bound on the indices of the generators possibly occurring in the expressions of an element.

Lemma 3.1.Define theceilingdweof a nonempty wordw=θi₁· · ·θi_` of Θ^∗ by (9) dwe:= max{ip+`−p|p= 1, . . . , `}.

Thendweis invariant under≡.

Proof. It suffices to consider the case of two wordsw, w⁰deduced from one another by applying one relation ofPH. Forw=uθjθiv andw⁰=uθiθj+1v, withj >i+ 2, one findsdwe= max(due+|v|+ 2, j+ 1 +|v|,dve) =dw⁰e. Similarly, forw=uθiθi+1θi+3v andw⁰=uθi+1θiθi+1v, one obtainsdwe= max(due+|v|+3, i+3+|v|,dve) =dw⁰e.

For a in H⁺, we write dae for the common value of dwe for w representing a.

A direct application is the following a priori nontrivial result:

Proposition 3.2.The word problem for PH is decidable.

Proof. For every wordwin Θ^∗, the≡-class ofwis finite: indeed,w⁰≡wimplies both

|w⁰|=|w|anddw⁰e=dwe, and the number of words w⁰ satisfying these conditions is bounded above by dwe^|w|. Therefore, starting from two wordsw, w⁰, one can decide whetherw⁰≡wholds by saturating{w}with respect to the relations ofPH, eventu- ally obtaining in finitely many steps an exhaustive enumeration of the≡-class ofw.

Then one comparesw⁰ with the elements of the list so constructed.

Another property that directly follows from the presentation is the fact that the monoidF⁺ is a quotient ofH⁺:

Proposition 3.3.The map π:θ_i7→τ_i induces a surjective homomorphism from the monoidH⁺ onto the Thompson monoid F⁺.

Proof. Letπ^∗ be the extension of πinto a homomorphism from the free monoid Θ^∗ to the monoid F⁺. We claim that w ≡w⁰ impliesπ^∗(w) = π^∗(w⁰). It is enough to check this whenw=w⁰ is a relation ofPH. The case of length 2 relations is trivial, as the latter are relations ofPF. For length 3 relations, we find in F⁺

π^∗(θi+1θiθi+1) =τi+1τiτi+1=τiτi+2τi+1=τiτi+1τi+3 =π^∗(θiθi+1θi+3).

Soπ^∗ induces a homomorphism from H⁺ to F⁺. The latter is surjective since each

generatorτi lies in the image.

Garside combinatorics for Thompson’s monoid <span class="mathjax-formula">$F^+$</span> and a hybrid with the braid monoid <span class="mathjax-formula">$B_{\infty }^{+}$</span>

ALGEBRAIC COMBINATORICS

Patrick Dehornoy & Emilie Tesson

Garside combinatorics for Thompson’s monoid F + and a hybrid with the braid

monoid B ∞ +

Garside combinatorics for Thompson’s monoid F ⁺ and a hybrid with the braid

monoid B _∞ ⁺