Patrick Tamari Dehornoy

Thompson Monoid 2011 Lattices

Hal

Thompson Monoid

TAMARI LATTICES AND THE SYMMETRIC

Keywords: 1991 Mathematics Subject Classification. 06B10, 06F15, 20F38, 20M05 Tamari lattice, Thompson groups, group of fractions, least common multiple, greatest common divisor, subword reversing

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Introduction

The aim of this text is to show the interest of using monoid techniques to investigate Tamari lattices. More precisely, we shall describe the very close connection existing between Tamari lattices and a certain submonoid F + sym of Richard Thompson's group F : equipped with the left-divisibility relation, the monoid F + sym is a lattice that includes all Tamari lattices. Roughly speaking, the principle is to attribute to the edges of the Tamari lattices names that live in the monoid F + sym. By using the subword reversing method, a general technique from the theory of monoids, we then obtain a very simple way of reproving the existence of the lattice operations, computing them, and establishing further properties.

The existence of a connection between Tamari lattices, associativity, and the Thompson group F has been known for decades and belongs to folklore. What is specific here is the role of the monoid F + sym, which is especially suitable for formalizing the connection. Some of the results already appeared, implicitly in [START_REF] Dehornoy | The structure group for the associativity identity[END_REF] and explicitly in [START_REF]Geometric presentations of Thompson's groups[END_REF]. Several new results are established in the current text, in particular the construction of a unique normal form in the monoid F + sym and the group F (Subsection 4.4) and the (surprising) result that the embedding of the monoid F + sym in the Thompson group F is not a quasiisometry (Proposition 5.8). In the language of binary trees, this implies that, for every constant C, there exist chains of length ℓ whose endpoints can be connected by a path of length at most ℓ/C (Corollary 5.13).

Let us mention that a connection between the Tamari lattices and the group F is described in [START_REF] Šunić | Tamari lattices, forests and Thompson monoids[END_REF]. However both the objects and the technical methods are disjoint from those developed below. In particular, the approach of [START_REF] Šunić | Tamari lattices, forests and Thompson monoids[END_REF] does not involve the symmetric monoid F + sym, which is central here, but it uses instead the standard Thompson monoid F + , which is not directly connected with the Tamari ordering.

The text is organized as follows. In Section 2, we recall the definition of Tamari lattices and Thompson's group F , and we establish a presentation of F in terms of some specific, non-standard generators a α indexed by binary addresses. In Section 3, we investigate the submonoid F + sym of F generated by the elements a α , we prove that F + sym equipped with divisibility has the structure of a lattice, and we describe the (close) connection between this lattice and Tamari lattices. Then, in Section 4, we use the Polish encoding of trees to construct an algorithm that computes common upper bounds for trees in the Tamari ordering and we deduce a unique normal form for the elements of F and F + sym. Finally, in Section 5, we gather results about the length of the elements of F with respect to the generators a α or, equivalently, about the distance in Tamari lattices, with a specific interest on lower bounds.

The framework

The aim of this section is to set our notation and basic definitions. In Subsections 2.1 and 2.2, we briefly recall the definition of the Tamari lattices in terms of parenthesized expressions and of binary trees, whereas Subsections 2.3 and 2.4 contain an introduction to Richard Thompson's group F and its action by rotation on trees. This leads us naturally to introducing in Subsection 2.5 a new family of generators of F indexed by binary addresses, and giving in Subsection 2.6 a presentation of F in terms of these generators.

2.1. Parenthesized expressions and associativity. Introduced by Dov Tamari in his 1951 PhD thesis, and appearing in the 1962 article [START_REF] Tamari | The algebra of bracketings and their enumeration[END_REF]-also see [START_REF] Friedman | Problèmes d'associativité : Une structure de treillis finis induite par une loi demi-associative[END_REF] and [START_REF] Huang | Problems of associativity: A simple proof for the lattice property of systems ordered by a semi-associative law[END_REF]the nth Tamari lattice, here denoted by T n , is, for every positive integer n, the poset (partially ordered set) obtained by considering all well-formed parenthesized expressions involving n + 1 fixed variables and declaring that an expression E is smaller than another one E ′ , written E T E ′ , if E ′ may be obtained from E by applying the associative law x(yz) = (xy)z in the left-to-right direction. As established in [START_REF] Tamari | The algebra of bracketings and their enumeration[END_REF], the poset (T n , T ) is a lattice, that is, any two elements admit a least upper bound and a greatest lower bound. Moreover, (T n , T ) admits a top element, namely the expression in which all left parentheses are gathered on the left, and a bottom element, namely the expression in with all right parentheses are gathered on the right.

As associativity does not change the order of variables, we may forget about their names, and use • everywhere. So, for instance, there exist five parenthesized expressions involving four variables, namely

•(•(••)), •((••)•), (•(••))•, (••)(••),

and ((••)•)•, and we have •((••)•) T (•(••))

• in the Tamari order as one goes from the first expression to the second by applying the associativity law with x = •, y = ••, and z = •. The Hasse diagrams of the lattices T 3 and T 4 respectively are the pentagon and the 14 vertex polyhedron displayed in Figures 1 and4 below. As is well known, the number of elements of T n is the nth Catalan number 1 n+1 2n n . The Tamari lattice T n is connected with a number of usual objects. For instance, its Hasse diagram is the 1-skeleton-that is, the graph made of the 0-and 1-cells-of the nth Mac Lane-Stasheff associahedron [START_REF] Lane | Natural associativity and commutativity[END_REF][START_REF] Stasheff | Homotopy associativity of H-spaces[END_REF]. Also T n embeds in the lattice made by the symmetric group S n equipped with the weak order: T n identifies with the sub-poset made by all 312-avoiding permutations (Björner & Wachs [START_REF] Björner | Shellable nonpure complexes and posets. II[END_REF]).

•(•(••)) •((••)•) (•(••))• ((••)•)• (••)(••)
For every n, replacing in a parenthesized expression the last (rightmost) symbol • with •• defines an embedding ι n of T n into T n+1 . We denote by T ∞ the limit of the direct system (T n , ι n ) so obtained. Note that T ∞ has a bottom element, namely the class of •, which is also that of ••, •(••), •(•(••)), etc., but no top element.

Trees and rotations.

There exists an obvious one-to-one correspondence between parenthesized expressions involving n + 1 variables and size n binary rooted trees, that is, trees with n interior nodes and n + 1 leaves, see Figure 2. In this text, we shall use both frameworks equivalently. We denote by T 0 ∧ T 1 the tree whose left-subtree is T 0 and whose right-subtree is T 1 , but skip the symbol ∧ in concrete examples involving •.

• •• (••)• •(••) •((••)•) Figure 2.
Correspondence between parenthesized expressions and trees.

When translated in terms of trees, the operation of applying associativity in the leftto-right direction corresponds to performing one left-rotation, namely replacing some subtree of the form T 0 ∧ (T 1 ∧ T 2 ) with the corresponding tree (T 0 ∧ T 1 ) ∧ T 2 , see Figure 3. So the Tamari lattice T n is also the poset of size n trees ordered by the transitive closure of left-rotation. We naturally use T for the latter partial ordering.

T 0 T 1 T 2 T 0 T 1 T 2 α α
Figure 3. Applying a left rotation in a tree: replacing some subtree of the form T0 ∧ (T1 ∧ T2) with the corresponding tree (T0 ∧ T1) ∧ T2.

In terms of trees, the bottom element of the Tamari lattice T n is the size n right-comb (or right-vine) C n recursively defined by C 0 = • and C n = • ∧ C n-1 for n 1, whereas the top element is the size n left-comb (or left-vine) C n recursively defined by

C 0 = • and C n = C n-1 ∧ • for n 1.
2.3. Richard Thompson's group F . Introduced by Richard Thompson in 1965, the group F appeared in print only later, in [START_REF] Mckenzie | An elementary construction of unsolvable word problems in group theory[END_REF] and [START_REF] Thompson | Embeddings into finitely generated simple groups which preserve the word problem[END_REF]. The most common approach is to define F as a group of piecewise linear self-homeomorphisms of the unit interval [0, 1].

Definition 1. The Thompson group F is the group of all dyadic order-preserving selfhomeomorphisms of [0, 1], where a homeomorphism f is called dyadic if it is piecewise linear with only finitely many breakpoints, every breakpoint of f has dyadic rational coordinates, and every slope of f is an integral power of 2. Typical elements of F are displayed in Figure 5. In this paper, it is convenient to equip F with reversed composition, that is, f g stands for f followed by g-using the other convention simply amounts to reversing all expressions. The notation x 0 is traditional for the element of F defined by

•(•(•(••))) •((••)(••)) •(•((••)•)) •((•(••))•) (•(•(••)))• •(((••)•)•) (•((••)•))• ((••)(••))• (••)((••)•) ((•(••))•)• (((••)•)•)•) (••)(•(••)) ((••)•)(••) (•(••))(••)
x 0 (t) =      t 2 for 0 t 1 2 , t -1 4 for 1 2 t 3 4 , 2t -1 for 3 4 t 1,
and x i is used for the element that is the identity on [0, 1 -1 2 i ] and is a rescaled copy of x 0 on [1 -1 2 i , 1]-see Figure 5 again. It is easy to check that F is generated by the sequence of all elements x i , with the presentation (2.1)

x 0 , x 1 , ... | x n+1 x i = x i x n for i < n .
One deduces that F is also generated by x 0 and x 1 , with the (finite) presentation

(2.2) x 0 , x 1 | [x -1 0 x 1 , x 0 x 1 x -1 0 ], [x -1 0 x 1 , x 2 0 x 1 x -2 0 ] , where [x, y] denotes the commutator xyx -1 y -1 .
The group F has many interesting algebraic and geometric properties, see [START_REF] Cannon | Introductory notes on Richard Thompson's groups[END_REF]. Its center is trivial, the derived group [F, F ] is a simple group, F includes no free group of rank more than 1 (Brin-Squier [START_REF] Brin | Groups of piecewise linear homeomorphisms of the real line[END_REF]), its Dehn function is quadratic (Guba [17]). It is 0 1

x 0 : 0 1

x 1 :

x 0 : x 1 : not known whether F is automatic, nor whether F is amenable. The latter question has received lot of attention as F seems to lie very close to the border between amenability and non-amenability.

Owing to the developments of Section 3 below, we mention one more (simple) algebraic result, namely that F is a group of (left)-fractions, that is, there exists a submonoid of F such that every element of F can be expressed as f -1 g with f, g in the considered submonoid.

Proposition 2.1. [START_REF] Cannon | Introductory notes on Richard Thompson's groups[END_REF] Define the Thompson monoid F + to be the submonoid of F generated by the elements x i with i 1. Then, as a monoid, F + admits the presentation (2.1), and F is a group of left-fractions for F + .

Thus F + consists of the elements of F that admit at least one expression in terms of the elements x i in which no factor x -1 i occurs. Although easy, Proposition 2.1 is technically significant as its leads to a unique normal form for the elements of F .

2.4.

The action of F on trees. An element of F is determined by a pair of dyadic decompositions of the interval [0, 1] specifying the intervals on which the slope has a certain value, and, from there, by a pair of trees.

To make the description precise, define a dyadic decomposition of [0, 1] to be an increasing sequence (t 0 , ..., t n ) of dyadic numbers with t 0 = 0 and t n = 1, such that no interval [t i , t i+1 ] may contain a dyadic number with denominator less that those of t i and t i+1 : for instance, (0, 1 2 , 3 4 , 1) is legal, but (0, 3 4 , 1) is not. Then dyadic decompositions are in one-to-one correspondence with binary rooted trees: the decomposition associated with • is (0, 1), whereas the one associated with T 0 ∧ T 1 is the concatenation of those associated with T 0 and T 1 rescaled to fit in [0, 1 2 ] and [ 1 2 , 1]. As the diagram representation of Figure 5 shows, every element of the group F is entirely specified by a pair of dyadic decompositions, hence by a pair of trees. Provided adjacent intervals are gathered, this pair of decompositions (hence of trees) is unique. We shall denote by (f -, f + ) the pair of trees associated with f . For instance, we have 1 -= 1 + = •, and, as illustrated in Figure 6,(x

0 ) -= •(••), (x 0 ) + = (••)•, (x 1 ) -= •(•(••)), and (x 1 ) + = •((••)•).
By construction, the trees f -and f + have the same size. Moreover, we have (f -1 ) -= f + and (f -1 ) + = f -as taking the inverse amounts to exchanging source and target in the diagram.

We now define a partial action of the group F on finite trees. Hereafter, we denote by B the family of all finite, binary, rooted trees, and by B # the family of all (finite, binary, rooted) labeled trees whose leaves wear labels in N. Thus B identifies with the family of all parenthesized expressions involving the single variable •, and B # with the family of all parenthesized expressions involving variables from the list {• 0 , • 1 , ...}. Forgetting the labels (or the indices of variables) defines a projection of B # onto B; by identifying • with • 0 , we can see B as a subset of B # . If T is a tree of B, we denote by T # the tree of B # obtained by attaching to the leaves of T labels 0, 1, ... starting from the left. Definition 2. A substitution is a map from N to B # . If σ is a substitution and T is a tree in B # , we define T σ to be the tree obtained from T by replacing every i-labeled leaf of T by the tree σ(i).

Formally, T σ is recursively defined by the rules

(• i ) σ = σ(i), (T 0 ∧ T 1 ) σ = T σ 0 ∧ T σ 1 . For instance, if T is • 3 (• 0 • 2 ) and we have σ(0) = •• and σ(2) = σ(3) = •, then T σ is •((••)•).
Definition 3. If T, T ′ are labeled trees and f is an element of the Thompson group F , we say that

T * f = T ′ holds if we have T = (f # -) σ and T ′ = (f # + ) σ for some substitution σ. Example 2.2. First consider f = 1. Then we have 1 -= 1 + = •, whence 1 # -= 1 # + = • 0 .
For every tree T , we have T = (1 # -) σ for any substitution satisfying σ(0) = T , and, in this case, we have (1 # -) σ = T . So T * 1 is always defined and it is equal to T . Consider now x 0 . Then we have

x 0-= •(••), whence x 0 # -= • 0 (• 1 • 2 )
. For a tree T , there exists a substitution satisfying T = (• 0 (• 1 • 2 )) σ if and only if T can be expressed as T 0 ∧ (T 1 ∧ T 2 ). In this case, the tree ((

• 0 • 1 )• 2 ) σ is (T 0 ∧ T 1 ) ∧ T 2 .
So T * x 0 is defined if and only if T is eligible for a left-rotation and, in this case, T * x 0 is the tree obtained from T by that left-rotation, see Figure 3.

Consider finally x 1 . Arguing as above, we see that T * x 1 is defined if and only if T can be expressed as

T 0 ∧ (T 1 ∧ (T 2 ∧ T 3 )), in which case T * f is the tree T 0 ∧ ((T 1 ∧ T 2 ) ∧ T 3 )
, that is, the tree obtained from T by a left-rotation at the right-child of the root. The above definition specifies what can naturally be called a partial action of the group F on (labeled) trees-labels are not important here as rotations do not change their order or repeat them, but they are needed for a clean definition of substitutions. Proposition 2.3. (i) For every (labeled) tree T and and every element f of F , there exists at most one T ′ satisfying T ′ = T * f .

T 0 T 1 T 2 T 0 T 1 T 2 T T * x 0 x 0 T 0 T 1 T 2 T 3 T 0 T 1 T 2 T 3 T T * x 1 x 1
(ii) For every (labeled) tree T , we have T * 1 = T .

(iii) For every (labeled) tree T and all f, g in F , we have (T * f ) * g = T * f g, this meaning that either both terms are defined and they are equal, or neither is defined. Moreover, for all f 1 , ..., f n in F , there exists T such that T * f i is defined for each i. Sketch, see [START_REF]Geometric presentations of Thompson's groups[END_REF] for details. (i) For f in F , a given tree T can be expressed in at most one way as (f # -) σ and, as the same variables occur on both sides of the associativity law, there is in turn at most one corresponding tree (f # + ) σ . Point (ii) has been established in Example 2.2. For (iii), the point is that there exists a simple rule for determining the pair of trees associated with f g. Indeed, we have (f g) -= f σ -and (f g) + = g τ + , where σ and τ are minimal substitutions satisfying

f σ + = g τ --that is, (σ, τ
) is a minimal identifier for f + and g -. As for the final point, it comes from the fact that, by construction, every tree f # has pairwise distinct labels and, therefore, a tree T can be expressed as (f # -) σ if and only if the skeleton of T (as defined in Definition 4 below) includes the skeleton of f # -. Then, for f 1 , ..., f n in F , one can always find a tree T whose skeleton includes those of (f 1 ) # -, ..., (f n ) # -. Proposition 2.4. For all (labeled) trees T, T ′ , the following are equivalent:

(i) One can go from T to T ′ using a finite sequence of rotations-that is, by applying associativity;

(ii) The trees T and T ′ have the same size, and the left-to-right enumerations of the labels in T and T ′ coincide;

(iii) There exists f in F satisfying T ′ = T * f . In this case, the element f involved in (iii) is unique. Sketch, see [START_REF]Geometric presentations of Thompson's groups[END_REF] for details. The equivalence of (i) and (ii) follows from the syntactic properties of the terms occurring in the associativity law, namely that the same variables occur on both sides, in the same order.

Next, assume that T and T ′ are equal size trees. Then T and T ′ determine dyadic decompositions of [0, 1], and there exists a dyadic homeomorphism f , hence an element of F , that maps the first onto the second. Provided the enumerations of the labels in T and T ′ coincide, we have T ′ = T * f . So (ii) implies (iii).

Conversely, we saw in Example 2.2 that the action of x 0 and x 1 is a rotation. On the other hand, we know that x 0 and x 1 generate F . Therefore, the action of an arbitrary element of f is a finite product of rotations. So (iii) implies (ii).

Finally, the uniqueness of the element f possibly satisfying T ′ = T * f follows from the fact that the pair (T, T ′ ) determines a unique pair of dyadic decompositions of [0, 1], so it directly determines the graph of the dyadic homeomorphism f . Proposition 2.4 states that F is the geometry group of associativity in the sense of [START_REF]Geometric presentations of Thompson's groups[END_REF]. A similar approach can be developed for every algebraic law, and more generally every family of algebraic laws, leading to a similar geometry monoid (a group in good cases). In the case of associativity together with commutativity, the geometry group happens to be the Thompson group V , whereas, in the case of the left self-distributivity law x(yz) = (xy)(xz), the geometry group is a certain ramified extension of Artin's braid group B ∞ [START_REF]Braids and Self-Distributivity[END_REF]-also see the case of x(yz) = (xy)(yz) in [START_REF]Study of an identity[END_REF]. In the latter cases, the situation is more complicated than with associativity as, in particular, the counterparts of (i) and (ii) in Proposition 2.4 fail to be equivalent.

2.5.

The generators a α . Considering the action of the group F on trees invites us to introducing, beside the standard generators x i , a new, more symmetric family of generators for F .

In order to define these elements, we need an index system for the subtrees of a tree. A common solution consists in describing the path connecting the root of the tree to the root of the considered subtree using (for instance) 0 for "forking to the left" and 1 for "forking to the right". Definition 4. A finite sequence of 0's and 1's is called an address; the empty address is denoted by ∅. For T a tree and α a short enough address, the α-subtree of T is the part of T that lies below α. The set of all α's for which the α-subtree of T exists is called the skeleton of T .

Formally, the α-subtree is defined by the following recursive rules: the ∅-subtree of T is T , and, for α = 0β (resp. 1β), the α-subtree of T is the β-subtree of T 0 (resp. T 1 ) if T is T 0 ∧ T 1 , and it is undefined otherwise.

Example 2.5. For T = •((••)•) (the rightmost example in Figure 2), the 10-subtree of T is ••, while the 01-and 111-subtrees are undefined. The skeleton of T consists of the seven addresses ∅, 0, 1, 10, 100, 101, and 11.

By definition, applying associativity in a parenthesized expression or, equivalently, applying a rotation in a tree T consists in choosing an address α in the skeleton of T and either replacing the α-subtree of T , supposed to have the form T 0 ∧ (T 1 ∧ T 2 ), by the corresponding (T 0 ∧ T 1 ) ∧ T 2 , or vice versa, see Figure 3 again. By Proposition 2.4, this rotation corresponds to a (unique) element of F . Definition 5. For every address α, we denote by a α the element of F whose action is a left-rotation at α. We denote by A A A the family of all elements a α for α an address.

According to Example 2.2 and Figure 7, the action of x 0 is a left-rotation at the root of the tree, and, therefore, we have x 0 = a ∅ . Similarly, x 1 is left-rotation at the rightchild of the root, that is, at the node with address 1, and, therefore, we have x 1 = a 1 . More generally, all elements a α can be expressed in terms of the generators x i , as will be done in Subsection 2.6 below. For the moment, we simply note that iterating the argument for x 1 gives for every i 1 the equality x i = a 1 i-1 , where 1 i-1 denotes 11...1, i -1 times 1.

The trees T such that T * a α is defined are easily characterized. Indeed, a necessary and sufficient for T * a α to exist is that the α-subtree of T is defined and a left-rotation can be applied to that subtree, that is, it can be expressed as

T 0 ∧ (T 1 ∧ T 2 )
. This is true if and only if the addresses α0, α10, and α11 lie in the skeleton of T , hence actually if and only if α10 lies in the skeleton of T since β0 may lie in the skeleton of a tree only if β1 and β do. Symmetrically, T * a -1 α is defined if and only if α01 lies in the skeleton of T . As a tree has a finite skeleton, there exist for every tree T finitely many addresses α such that T * a ±1 α is defined, see Figure 8. Before proceeding, we note that the forking nature of the family A A A naturally gives rise to a large family of shift endomorphisms of the group F . Lemma 2.6. For every address α, there exists a (unique) shift endomorphism sh α of F that maps a β to a αβ for every β.

Proof. For f in F , let sh 1 (f ) denote the homeomorphism obtained by rescaling f , applying it in the interval [ 1 2 , 1], and completing with the identity on [0, 1 2 ]. Then sh 1 is an endomorphism of F , and it maps x i to x i+1 for every i. Moreover, for every β, the

a -1 10 a -1 1 a 1 a ∅ ∅ 1 10 Figure 8. The four elements a ±1 α such that T * a ±1 α is defined in the case T = •(((••)•)(••)). element sh 1 (a β ) is the rescaled version of a β applied in the interval [ 1 2 , 1]. By definition, this is a 1β .
Symmetrically, for f in F , let sh 0 (f ) denote the homeomorphism obtained by rescaling f , applying it in the interval [0, 1 2 ], and completing with the identity on [ 1 2 , 1]. Then sh 0 is an endomorphism of F , and, for every β, the element sh 0 (a β ) is the rescaled version of a β applied in the interval [0, 1 2 ], hence it is a 0β . Finally, we recursively define sh α for every α by sh ∅ = id F and, for i = 0, 1, sh iα (f ) = sh i (sh α (f )). By construction, sh α (a β ) = a αβ holds for all α, β.

2.6. Presentation of F in terms of the elements a α . As the family {x 0 , x 1 }, which is {a ∅ , a 1 }, generates the group F , the family A A A generates F as well. By using the presentation (2.1) or (2.2), we could easily deduce a presentation of F in terms of the elements a α . However, we can obtain a more natural and symmetric presentation by coming back to trees and associativity, and exploiting the geometric meaning of the elements of A A A.

Lemma 2.7. Say that two addresses α, β are orthogonal, written α ⊥ β, if there exists γ such that α begins with γ0 and β begins with γ1, or vice versa. Then all relations of the following family R R R are satisfied in F :

a α a β = a β a α for α ⊥ β, (2.3) a α11β a α = a α a α1β , a α10β a α = a α a α01β , a α0β a α = a α a α00β , (2.4) a 2 α = a α1 a α a α0 . (2.5)
Proof. By Proposition 2.4, in order to prove that two elements f, f ′ of F coincide, it is enough to exhibit a tree T such that T * f and T * f ′ are defined and equal.

The commutation relations of type (2.3) are trivial. If α and β are orthogonal, the αand β-subtrees are disjoint, and the result of applying rotations (as well as any transformations) in each of these subtrees does not depend on the order. So we have sh α (f ) sh β (g) = sh β (g) sh α (f ) for all transformations f, g and, in particular, a α a β = a β a α .

The quasi-commutation relations of type (2.4) are more interesting. Assume that T, T ′ are trees and a ∅ maps T to T ′ . Then, by definition, the 1-subtree of T ′ is a copy of the 11-subtree of T . Now, assume that f is a (partial) mapping of B to itself. Then, starting from T , first applying a ∅ and then applying f to the 11-subtree leads to the same result as first applying f to the 1-subtree and then applying a ∅ , see Figure 9. Moreover, if f is a partial mapping, the result of one operation is defined if and only if the result of the other is. So, in all cases, we have

a ∅ sh 11 (f ) = sh 1 (f ) a ∅ .
Applying this to f = a β then gives a ∅ a 11β = a 1β a ∅ . Shifting by α this relation, we obtain a α a α11β = a α1β a α , the first relation of (2.4). Arguing similarly with the 0-and 10-subtrees in place of the 11-subtree, one obtains the other relations of (2.4).

Finally, the relations of (2.5) stem from the pentagon of Figure 1. As Figure 10 shows, the relation a 2 ∅ = a 1 a ∅ a 0 is satisfied in F and, therefore, so is its shifted version a 2 α = a α1 a α a α0 for every address α. It is then easy to check that the above relations actually exhaust the relations connecting the elements a α in the group F . Proposition 2.8. [START_REF] Dehornoy | The structure group for the associativity identity[END_REF][START_REF]Geometric presentations of Thompson's groups[END_REF] The group F admits the presentation

T T ′ f sh 11 (f ) sh 1 (f ) a ∅ a ∅ a ∅ a ∅ a 11 a 1
A A A | R R R .
Proof. By Lemma 2.7, the relations of R R R are valid in F . Conversely, to prove that these relations make a presentation, it is sufficient to show that they include the relations of a previously known presentation. This is what happens as, for 1 i < n, the relation

a 1 n a 1 i-1 = a 1 i-1 a 1 n-1 , which is a reformulation of the relation x n+1 x i = x i x n of (2.1), occurs in R R R as the first relation of (2.4) with α = 1 i-1 and β = 1 n-i .
As an application, we compute the elements a α in terms of the generators x i . Proposition 2.9. If α is an address containing at least one 0, say α = 1 i 0 1+i0 10 i1 ...10 im with m 0 and i, i 0 , ..., i m 0, then, putting g = x im+1 i+m+1 ... x i1+1 i+2 x i0+1 i+1 , we have

(2.6) a α = g -1 x -1 i+m+2 x i+m+1 g. Proof.
It is sufficient to establish the formula in the case i = 0 as, then, applying sh i 1 gives the general case. We use induction on (m, i 0 ) with respect to the lexicographical (well)-order, that is, (m ′ , i ′ 0 ) is smaller than (m, i 0 ) if and only if we have either m ′ < m, or m ′ = m and i ′ 0 < i 0 .

Assume first (m, i 0 ) = (0, 0), that is, α = 0. Then the pentagon relation at ∅ gives

a α = a 0 = a -1 ∅ a -1 1 a 2 ∅ = x -1 1 (x -1 2 x 1 )
x 1 , which is the expected instance of (2.6). Assume now m 1 and i 0 = 0, that is α = 010 i1 ... 10 im . Then the quasi-commutation relation for ∅ and 01β gives (2.7)

a α = a 010 i 1 ... 10 im = a -1 ∅ a 10 1+i 1 10 i 2 ... 10 im a ∅ = x -1 1 (a 10 1+i 1 10 i 2 ... 10 im )x 1 .
The number of non-initial symbols 1 in 0 1+i1 10 i2 ... 10 im is m -1. As (m -1, i 1 ) is smaller than (m, 0), the induction hypothesis gives a 0 1+i 1 10 i 2 ...

10 im = g -1 x -1 m+1 x m g with g = x im+1 m ... x i2+1 2 x i1+1 1 .
Using sh 1 , we get a 10 1+i 1 10 i 2 ...

10 im = h -1 x -1 m+2 x m+1 h with h = x im+1 m+1 ... x i2+1 3 x i1+1 2
. Merging with (2.7), we deduce the expected value for a α . Assume finally i 0 1. Then the quasi-commutation relation for ∅ and 00β gives

a α = a 0 1+i 0 10 i 1 ... 10 im = a -1 ∅ a 0 i 0 10 i 1 ... 10 im a ∅ = x -1 1 a 0 i 0 10 i 1 ... 10 im x 1 .
The pair (m, i 0 -1) is smaller than the pair (m, i 0 ), so the induction hypothesis gives

a α = x -1 1 (g -1 x -1 m+2 x m+1 g)x 1 with g = x im+1 m+1 ... x i1+1 2
x i0 1 , again the expected instance of (2.6). So the induction is complete.

Example 2.10. Consider α = 01100, which corresponds to m = 2, and i = i 0 = i 1 = 0, and i 2 = 2. Then we find

a α = g -1 x -1 4 x 3 g with g = x 3 3 x 2 x 1 , that is, a 01100 is equal to x -1 1 x -1 2 x -3 3 x -1 4 x 4 3 x 2 x 1 .

A lattice structure on the Thompson group F

Here comes the core of our study, namely the investigation of the submonoid F + sym of F generated by the elements a α . The main result is that F + sym has the structure of a lattice when equipped with its divisibility relation, and that this lattice is closely connected with the Tamari lattices, which occur as initial sublattices.

These results are not trivial, as, in particular, determining a presentation of F + sym is not so easy. Our approach relies on using subword reversing, a general method of combinatorial group theory that turns out to be well suited for F + sym. One of the outcomes is a new proof (one more!) of the fact that Tamari posets are lattices.

The section is organized as follows. The symmetric Thompson monoid F + sym is introduced in Subsection 3.1, and it is investigated in Subsection 3.2 using subword reversing. The lattice structure on F + sym and its connection with the Tamari lattices are described in Subsection 3.3. Finally, a few results about the algorithmic complexity of the reversing process are gathered in Subsection 3.4.

The symmetric Thompson monoid

F + sym.
Once new generators a α of the Thompson group F have been introduced, it is natural to investigate the submonoid generated by these elements. Definition 6. The symmetric Thompson monoid F + sym is the submonoid of F generated by the elements a α with α a binary address.

The family A A A of all elements a α is a sort of closure of the family of standard generators x i under all local left-right symmetries, so the above terminology is natural. Another option could be to call F + sym the dual Thompson monoid as the relation of F + and F + sym is reminiscent of the relation of the standard braid monoids and the dual braid monoids generated by the Birman-Ko-Lee braids. Although straightforward, the following connection is essential for our purpose:

Lemma 3.1. For all trees T, T ′ , the following are equivalent (i) We have T T T ′ in the Tamari order;

(ii) There exists f in F + sym satisfying T ′ = T * f . Proof. By definition, T T T ′ holds if there exists a finite sequence of left-rotations transforming T into T ′ . Now applying the left-rotation at α is letting a α act.

In order to investigate the monoid F + sym and its connection with the Tamari lattices, it will be necessary to first know a presentation of F + sym. Owing to Propositions 2.1 and 2.8, the following result should not be a surprise. Proposition 3.2. [START_REF] Dehornoy | The structure group for the associativity identity[END_REF][START_REF]Geometric presentations of Thompson's groups[END_REF] The monoid F + sym admits the presentation A A A | R R R + , and F is a group of right-fractions for F + sym (that is, every element of F can be expressed as f g -1 with f, g in F + sym).

However, the proof of Proposition 3.2 is more delicate than the proof of Proposition 2.1, and no very simple argument is known.

Sketch of the proof developed in [START_REF] Dehornoy | The structure group for the associativity identity[END_REF][START_REF]Geometric presentations of Thompson's groups[END_REF]. In order to prove that the relations of R R R generate all relations connecting the elements a α in the monoid F + , one introduces, for every size n tree T , an explicit sequence c T of elements a α satisfying C n * c T = T -as will be made in the proof of Proposition 3.9 below. The point is then to show that, if T ′ = T * w holds, then the relations of R R R are sufficient to establish the equivalence of c T ′ and c T w. Then, if two A A A-words u, v represent the same element of F + sym, and T is a tree such that both T * u and T * v are defined, the above argument shows that c T u and

c T v are R R R- equivalent, since both are R R R-equivalent to c T * u . Provided R R R-equivalence is known to allow left-cancellation, one deduces that u and v are R R R-equivalent, as expected.
Here we shall propose a new proof, which is more lattice-theoretic in that it exclusively relies on the so-called subword reversing method, which we shall see below is directly connected with the Tamari lattice operations. Instead of working with F + sym, we investigate the abstract monoid A A A | R R R + defined by the presentation (A A A, R R R) of Proposition 2.8. A priori, as F + sym is generated by A A A and satisfies the relations of R R R, we only know that

F + sym is a quotient of A A A | R R R + .
Definition 7. Assume that M is a monoid. For f, g in M , we say that f left-divides g, or that g is a right-multiple of f , written f g, if f g ′ = g holds for some g ′ of M . We use Div(f ) for the family of all left-divisors of f . It is standard that the left-divisibility relation is a partial pre-ordering. Moreover, if M contains no invertible element except 1, this partial pre-ordering is a partial ordering, that is, the conjunction of f g and g f implies f = g. Lemma 3.3. In order to establish Proposition 3.2, it is sufficient to prove that the monoid A A A | R R R + is cancellative and any two elements admit a common right-multiple.

Proof. A classical result of Ore (see for instance [START_REF] Clifford | The algebraic theory of semigroups[END_REF]) says that, if a monoid M is cancellative and any two elements of M admit a common right-multiple, then M embeds in a group of right-fractions G. Moreover, if M admits the presentation A | R + , then G admits the presentation A | R . So, if the hypotheses of the lemma are satisfied, then the monoid A A A | R R R + embeds in a group of fractions that admits the presentation

A A A | R R R . By Proposition 2.8, the group A A A | R R R is the group F . Therefore, A A A | R R R + is isomorphic to the submonoid of F generated by A A A, that is, to F + sym.
Hence F + sym admits the expected presentation, and F is a group of right-fractions for F + sym.

Subword reversing.

In order to apply the strategy of Lemma 3.3, we have to prove that the presented monoid A A A | R R R + is cancellative and any two elements of A A A | R R R + admit a common right-multiple. The subword reversing method [START_REF]Groups with a complemented presentation[END_REF][START_REF]Complete positive group presentations[END_REF] proves to be relevant. We recall below the basic notions, and refer to [START_REF]The word reversing method[END_REF] or [START_REF] Dehornoy | Garside Theory[END_REF]Section II.4] for a more complete description.

Hereafter, words in an alphabet A are called (positive) A-words, whereas words in the alphabet A∪A -1 , where A -1 consists of a copy a -1 for each letter a of A, are called signed A-words. We say that a group presentation (A, R) is positive if all relations in R have the form u = v where u and v are nonempty positive A-words. We denote by A | R + and by A | R the monoid and the group presented by (A, R), respectively, and we use ≡ + R (resp. ≡ R ) for the congruence on positive A-words (resp. on signed A-words) generated by R. Finally, for w a signed A-word, we denote by w the element of A | R represented by w, that is, the ≡ R -class of w. Definition 8. Assume that (A, R) is a positive presentation. If w, w ′ are signed A-words, we say that w is right-R-reversible to w ′ in one step if w ′ is obtained from w either by deleting some length 2 subword a -1 a or by replacing some length 2 subword a -1 b with a word vu -1 such that av = bu is a relation of R. We write w R w ′ if w is right-Rreversible to w ′ in finitely many steps.

The principle of right-R-reversing is to use the relations of R to push the negative letters (those with exponent -1) to the right, and the positive letters (those with exponent +1) to the left. The process can be visualized in diagrams as in Figure 11.

Example 3.4. Consider the presentation (

A A A, R R R), which is positive. Let w be the signed A A A-word a -1
1 a ∅ a -1 00 a 1 . Then w contains two negative-positive length 2 subwords, namely a -1 1 a ∅ and a -1 00 a 1 . There exists in R R R a unique relation of the form a 1 ... = a ∅ ... , namely a 1 a ∅ a 0 = a 2 ∅ , and a unique relation a 00 ... = a 1 ... , namely a 00 a 1 = a 1 a 00 . Therefore, there exists two ways to right-R R R-reverse w, namely replacing a -1 1 a ∅ with a ∅ a -1 00 a -1 ∅ and obtaining w 1 = a ∅ a 0 a -1 ∅ a 1 , or replacing a -1 00 a 1 with a 1 a -1 00 and obtaining w ′ 1 = a -1 1 a ∅ a 1 a -1 00 . The words w 1 and w ′ 1 each contain a unique negative-positive length 2 subword, and reversing it leads in both cases to w 2 = a ∅ a 0 a -1 ∅ a 1 a -1 00 . The word w 2 contains a unique negative-positive length two subword and reversing it leads to w 3 = a ∅ a 0 a ∅ a -1 0 a -1 ∅ a -1 00 . As the latter word contains no negative-positive subword, no further right-reversing is possible. See Figure 11.

It is easy to see that, if (A, R) is a positive presentation and w, w ′ are signed A-

words, then w R w ′ implies w ≡ R w ′ and that, if u, v, u ′ , v ′ are positive A-words, then u -1 v R v ′ u ′-1 implies uv ′ ≡ + R vu ′ .
In particular, using ε for the empty word,

(3.1) u -1 v R ε implies u ≡ + R v.
In general, (3.1) need not be an equivalence, but it turns out that this is the interesting situation, in which case the presentation (A, R) is said to be complete with respect to right-reversing. Roughly speaking, a presentation is complete with respect to rightreversing if right-reversing always detects equivalence. The important point here is that the presentation (A A A, R R R) has this property.

Lemma 3.5. [START_REF] Dehornoy | The structure group for the associativity identity[END_REF] The presentation (A A A, R R R) is complete with respect to right-reversing.

a 1 a ∅ a ∅ a 0 a ∅ a ∅ a 0 a 00 a ∅ a 1 a 1 a 00 Figure 11. Right-R R R-reversing of the signed A A A-word a -1
1 a ∅ a -1 00 a1: we draw the initial word as a zigzag path (here in dark green) from SW to NE by associating with every letter a a horizontal arrow labeled a and every letter a -1 a vertical arrow labeled a (crossed in the wrong direction); then reversing a -1 b to vu -1 corresponds to closing the open pattern made by a vertical aarrow and a horizontal b-arrow with the same source by adding horizontal arrows labeled v and vertical arrows labeled u; the final word corresponds to the rightmost path from the SW corner to the NE corner, here a ∅ a0a ∅ a -1 0 a -1 ∅ a -1 00 (light green).

Sketch. By [13, Proposition 2.9], a sufficient condition for a positive presentation (A, R) to be complete with respect to right-reversing is that (A, R) satisfies

(3.

2)

(i) There exists a ≡ + R -invariant map λ from positive A-words to N satisfying λ(uv) λ(u) + λ(v) for all u, v and λ(a) 1 for a in A, and (ii) For all a, b, c in A and all positive A-words u, v, if a -1 cc -1 b R vu -1 holds, then v -1 a -1 bu R ε holds as well.

We claim that (A A A, R R R) satisfies (3.2). As for (i), we cannot use for λ the length of words, as it is not ≡ + R R R -invariant: in the pentagon relation, the length 2 word a 2 ∅ is ≡ + R R R -equivalent to the length 3 word a 1 a ∅ a 0 . Now, for T a tree, let µ(T ) be the total number of 0's occurring in the addresses of the leaves of T : for instance, we have µ(•(••)) = 2 and µ((••)•) = 3, as the leaves of •(••) have addresses 0, 10, 11, with two 0's, and those of (••)• have addresses 00, 01, 1, with three 0's. Then put

(3.3) λ(w) = µ(w + ) -µ(w -).
For instance, if w is a ∅ , the trees w -and w + are •(••) and (••)•, and one finds λ(a ∅ ) = 3 -2 = 1. A similar argument gives λ(a α ) = 1 for every address α. More generally, one easily checks that T T T ′ implies µ(T ′ ) µ(T ). Hence the function λ takes values in N. Moreover, a counting argument shows that, in the previous situation, µ(T ′σ )µ(T σ ) µ(T ′ )µ(T ) holds for every substitution σ. If u and v are positive A A A-words, then, as seen in the proof of Proposition 2.3, we have (uv) -= u σ -and (uv)

+ = v τ + for some substitutions σ, τ satisfying u σ + = v τ -. We deduce λ(uv) = µ(uv + ) -µ(uv -) = µ(v τ + ) -µ(u σ -) = µ(v τ + ) -µ(v τ -) + µ(u σ + ) -µ(u σ -) µ(v + ) -µ(v -) + µ(u + ) -µ(u -) = λ(v) + λ(u).
As for (ii), the problem is to check that, whenever α, β, γ are addresses and the signed word a -1 α a γ a -1 γ a β is right-R R R-reversible to some positive-negative word vu -1 , then v -1 a -1 α a β u is right-R R R-reversible to the empty word. The systematic verification seems tedious. Actually it is not. First, what matters is the mutual position of the addresses α, β, γ with respect to the prefix ordering, and only finitely many patterns may occur. Next, for every pair of addresses α, β, there exists in R R R exactly one relation of the form a α ... = a β ... , which implies that, for every signed A A A-word w, there exists at most one pair of positive A A A-words u, v such that w is right-R R R-reversible to vu -1 . Finally, all instances involving quasi-commutation relations turn out to be automatically verified. So, the only critical cases are those corresponding to the triple of addresses ∅, 1, 11 and its translated and permuted copies, and a direct verification is then easy. For instance, the reader can see on Figure 12 that we have a

-1 ∅ a 1 a -1 1 a 11 R R R a 2 ∅ a -1 00 a -1 0 a -1 ∅ a -1 01 a -1 1 and a -3 ∅ a 11 a 1 a 10 a ∅ a 0 a 00 R R R ε. ∅ 1 0 1 ∅ 0 01 ∅ ∅ ∅ 0 ∅ 0 ∅ 0 00 1 11 ∅ 11 1 10 ∅ 0 00 1 ∅ ∅ ∅ ∅ ∅ 00 0 ∅ 10 1 ε ε ε ε ε ε Figure 12. Proof of Lemma 3.5: a -1 ∅ a1a -1 1 a11 is right-R R R-reversible to a 2 ∅ a -1 00 a -1 0 a -1 ∅ a -1 01 a -1 1
(left), and a -3 ∅ a11a1a10a ∅ a0a00 is right-R R Rreversible to the empty word; dotted lines represent the empty word that appears when a pattern a -1 α aα is reversed.

Once a positive presentation is known to be complete with respect to right-reversing, it is easy to deduce properties of the associated monoid.

Proposition 3.6. The monoid

A A A | R R R + is left-cancellative.
Proof. By [13, Proposition 3.1], if (A, R) is a positive presentation that is complete with respect to right-reversing, a sufficient condition for the monoid A | R + to be leftcancellative is that

(3.4)
R contains no relation of the form au = av with a in A and u = v.

By definition, R R R satisfies (3.4). Hence the monoid

A A A | R R R + is left-cancellative.
As for right-cancellation, no new computation is needed as we can exploit the symmetries of R R R. First, we introduce a counterpart of right-reversing where the roles of positive and negative letters are exchanged. Definition 9. Assume that (A, R) is a positive presentation. If w, w ′ are signed Awords, we say that w is left-R-reversible to w ′ in one step if w ′ is obtained from w either by deleting some length 2 subword aa -1 or by replacing some length 2 subword ab -1 with a word u -1 v such that ua = vb is a relation of R. We write w R w ′ if w is left-R-reversible to w ′ in finitely many steps.

Of course, properties of left-reversing are symmetric to those of right-reversing.

Proposition 3.7. The monoid

A A A | R R R + is right-cancellative.
Proof. The argument is symmetric to the one for Proposition 3.6, and relies on first proving that (A A A, R R R) is, in an obvious sense, complete with respect to left-reversing. Due to the symmetries of R R R, this is easy. Indeed, for w a signed A A A-word, let w denote the word obtained by reading the letters of w from right to left, and exchanging 0 and 1 everywhere in the indices of the letters a α . For instance, a 110 a ∅ is a ∅ a 001 . A direct inspection shows that the family R R R of all relations u = v for u = v a relation of R R R is R R R itself. It follows that, for all signed A A A-words w, w ′ , the relations w R R R w ′ and w R R R w ′ are equivalent. Then, as w → w is an alphabetical anti-automorphism, the completeness of (A A A, R R R) with respect to right-reversing implies the completeness of (A A A, R R R), hence of (A A A, R R R), with respect to leftreversing. As the right counterpart of (3.4) is satisfied, we deduce that the monoid

A A A | R R R + is right-cancellative.
In order to complete the proof of Proposition 3.2 using the strategy of Lemma 3.3, we still need to know that any two elements of the monoid A A A | R R R + admit a common right-multiple. Using the action on trees, it is easy to prove that result in F + sym. But this is not sufficient here as we do not know yet that F + sym is isomorphic to A A A | R R R + . We appeal to right-reversing once more.

Proposition 3.8. Any two elements of

A A A | R R R + admit a common right-multiple. Proof. If (A, R) is a positive presentation, saythat right-R-reversing is terminating if, for all positive A-words u, v, there exist positive A-words u ′ , v ′ satisfying u -1 v R R R v ′ u ′-1 .
We noted that the latter relation implies uv ′ ≡ + R vu ′ , thus implying that, in the monoid A | R + , the elements represented by u and v admit a common right-multiple. So, in order to establish the proposition, it is sufficient to prove that right-R R R-reversing is terminating, a non-trivial question as, because of the pentagon relations, the length of the words may increase under right-reversing, and there might exist infinite reversing sequences-try right-reversing of a -1 ba in the presentation (a, b, ab = b 2 a). Now, by [13, Proposition 3.11], if (A, R) is a positive presentation, a sufficient condition for right-R-reversing to be terminating is that (A, R) satisfies (3.5) (i) For all a, b in A, there is exactly one relation a ... = b ... in R, and (ii) There exists a family A of positive A-words that includes A and is closed under right-R-reversing, this meaning that, for all u, v in A, there exist

u ′ , v ′ in A # ∪ {ε} satisfying u -1 v R R R v ′ u ′-1 .
We claim that (A A A, R R R) satisfies (3.5). Indeed, (i) follows from an inspection of R R R. As for (ii), let us put âα,r = a α a α0 ... a α0 r-1

for α an address and r 1, see Figures 14 and19 for an illustration of the action of âα,r on trees. Then the family A A A of all words âα,r includes A A A as we have a α = âα,1 for every α, and it is closed under right-R R R-reversing as we find

(3.6) â-1 β,s âα,r R R R                    â0 s-r ,s-r for β = α with r < s, âα,r â-1 β,s for β ⊥ α, âα,r â-1 α0 r+1 γ,s for β = α0γ, âα,r â-1 α0 r 1γ,s for β = α10 r γ, âα,r â-1 α0 i 1γ,s
for β = α10 i 1γ with i < r, âα,r+s â-1

α0 i ,s
for β = α10 i with i < r, see Figure 13. Note that A A A is the smallest family that includes A A A and is closed under right-R R R-reversing as the last type of relation in (3.6) inductively forces any such family to contain âα,r for every r. In terms of the generators âα,r , the pentagon relation can be expressed as a 2 ∅ = a 1 â∅,2 , with both sides of length 2. The last type in (3.6) corresponds to an extended pentagon relation âα,r âα0 i ,s = âα10 i ,s âα,r+s for all r, s, i with i < r, whose counterpart in terms of tree rotation is displayed in Figure 14. We thus established that the monoid A A A | R R R + satisfies the conditions of Lemma 3.3 and, therefore, the proof of Proposition 3.2 is complete.

The lattice structure of F +

sym. Here comes the central point, namely the connection between the right-divisibility relation of the monoid F + sym, which we now know admits the presentation A A A | R R R + , and the Tamari posets. We recall that Div(f ) denotes the family of all left-divisors of f . Proposition 3.9. For every n 1, the subposet (Div(a n-1

∅

), ) of (F + sym,

) is isomorphic to the Tamari poset (T n , T ). The poset ( n Div(a n ∅ ), ) is isomorphic to the Tamari poset (T ∞ , T ).

Proof. An immediate induction gives the equality C n * a n-1 ∅ = C n for every n, that is, the element a n-1 ∅ of F + sym maps the right-comb C n to the left-comb C n . Hence C n * f is defined for every element f of F + sym that left-divides a n-1

∅

. Thus, as C n belongs to T n and the action of F + sym preserves the size of the trees, we obtain a well defined map (3.7)

I n : f → C n * f of Div(a n-1 ∅
) into T n . By Proposition 2.4, the map I n is injective. On the other hand, we claim that I n is surjective. To prove it, it suffices to exhibit, for every size n tree T , an element of F + sym that maps the right-comb C n to T . Now, for every tree T , define two elements c T , c ′ T of F + sym by the recursive rules:

(3.8) c T = 1 c ′ T0 sh 1 (c ′ T1 ) a ∅ c ′ T = 1 for T of size 0, c ′ T0 sh 1 (c ′ T1 ) a ∅ for T = T 0 ∧ T 1 .
For every size n tree T and every p ≥ 1, we have C n * c T = T and C n+p * c ′ T = T ∧ C p , as shows an induction on T : everything is obvious for T = •, and, for T = T 0 ∧ T 1 , it suffices to follow the diagrams of Figure 15. Note that introducing both c T and c ′ T is necessary for the induction. However, the connection c ′ T = c T a 1 i-1 ... a 1 a ∅ , where i is the length of the rightmost branch in T , is easy to check.

Thus I n is a bijection of Div(a n-1

∅

) onto T n . Moreover, I n is compatible with the orderings. Indeed, assume f g, say f g ′ = g. Then, by Proposition 2.3, we have (C n * f ) * g ′ = C n * g, whence C n * f T C n * g by Lemma 3.1. This completes the proof that (Div(a n-1 ∅ ), ) is isomorphic to the Tamari poset (T n , T ). As for T ∞ , we observe that, for every n, we have C n+1 = C # n σ where σ is the substitution that maps 0, ..., n -1 to • and n to ••. On the other hand, by definition, Div(a n-1

∅

) is an initial segment of Div(a n ∅ ) and, for every f in Div(a n-1

∅

), we have

C n+1 * f = C # n σ * f = (C n * f ) #σ , hence I n+1 (f ) = ι n (I n (f ))
. It follows that the family (I n ) n 1 induces a well defined map I ∞ of n Div(a n ∅ ) into T ∞ . The map I ∞ is injective because I n is, it is surjective as, by definition, T ∞ is the limit of the directed system (T n , ι n ), and it preserves the orderings as I n does. Remark 3.10. The subset n Div(a n ∅ ) involved in Proposition 3.9 is a proper subset of F + sym as, for instance, it contains no a α such that 0 occurs in α: indeed, in this case, C n * a α is not defined, whereas C n * f is defined for every f left-dividing a n-1 ∅ . The connection of Proposition 3.9 can be used in both directions. If we take for granted that the Tamari posets are lattices, we deduce that the subsets (Div(a n-1 ∅ ), ) of (F + sym,

C n C n1 T 0 T 1 T 0 c ′ T0 sh 1 (c T1 ) C n+p C n1+p C p C p T 0 T 0 T 1 T 0 T 1 c ′ T0 sh 1 (c ′ T1 ) a ∅
) must be lattices as well, that is, with the usual terminology of left-divisibility relation, that any two elements of n Div(a n ∅ ) admit a least common right-multiple, or right-lcm, and a greatest common left-divisor, or left-gcd.

On the other hand, if we have a direct proof that (F + sym,

) is a lattice, then the isomorphism of Proposition 3.9 provides a new proof of the lattice property for the Tamari posets. This is what happens. ) is a lattice. Corollary 3.12. For every n, the Tamari poset (T n , ) is a lattice.

To establish Proposition 3.11, we once again appeal to subword reversing. of Proposition 3.11. By [13, Proposition 3.6], if (A, R) is a positive presentation that is complete with respect to right-reversing, a sufficient condition for any two elements of A | R + that admit a common right-multiple to admit a right-lcm is that (A, R) satisfies Condition (i) of (3.5); moreover, in this case, the right-lcm of the elements represented by two A-words u, v is represented by uv ′ and vu ′ , where u ′ , v ′ are the positive A-words for which u -1 v R v ′ u ′-1 holds. Now, as already noted, (A A A, R R R) satisfies (3.5). Hence any two elements of F + sym that admit a common right-multiple admit a right-lcm. On the other hand, by Proposition 3.

8, any two elements of

A A A | R R R + , that is, of F +
sym, admit a common right-multiple. Hence any two elements of F + sym admit a right-lcm. In other words, any two elements in the poset (F + sym,

) admit a least upper bound. As for left-gcd's, we can argue as follows. Let denote the right-divisibility relation, which is the binary relation such that f g holds if and only if we have g ′ f = g for some g ′ (the difference with is that, here, f appears on the right and not on the left). Then we have the derived notions of a left-lcm and a right-gcd. An easy general result says that, if f, g, f ′ , g ′ are elements of a monoid and satisfy f g ′ = gf ′ , then f and g admit a left-gcd if and only if f ′ and g ′ admit a left-lcm. By Proposition 3.8, any two elements of F + sym admit a common right-multiple and so, it suffices to show that any two elements of F + sym admit a left-lcm to deduce that they admit a left-gcd. Now, the existence of left-lcm's in F + sym follows from the properties of left-R R R-reversing, which we have seen in the proof of Proposition 3.7 are similar to those of right-R R R-reversing. Remark 3.13. Another way of deducing the existence of left-gcd's from that of rightlcm's is to use Noetherianity properties. The existence of the function λ of (3.3) implies that a set Div(f ) contains no infinite increasing sequence

f 1 ≺ f 2 ≺ ... in F +
sym. For all f, g, the family Div(f ) ∩ Div(g) is nonempty as it contains 1, and, by Noetherianity, it contains a -maximal element, which must be a left-gcd of f and g.

3.4.

Computing the operations. We conclude this section with results about the algorithmic complexity of subword reversing in F + sym.

Here we concentrate on space complexity, namely bounds on the length of words; it would be easy to state analogous bounds on the number of reversing steps, hence for time complexity.

Proposition 3.14. If w, w ′ are signed A A A-words, w R R R w ′ implies |w ′ | |w| 2 /4 + |w|.
More precisely, we have |w ′ | p + q + pq if w contains p positive letters and q negative letters. These bounds are sharp.

Proof. By construction, the R R R-reversing steps in the right-R R R-reversing of w to w ′ can be gathered into R R R-reversing steps, which are at most pq in number. Consider the sum of the indices r of the involved generators âα,r . Each R R R-reversing step increases this sum by 1 at most (in the case of a pentagon relation), so the total sum in the final p + q generators âα,r is at most p + q + pq. So, when the generators âα,r are decomposed as products of a α 's, at most p + q + pq of the latter occur.

The bound is sharp, as an easy induction gives (a 1 p-1 ... a 1 a ∅ ) -1 a q 1 p R R R a q ∅ (â 1 p-1 ,q ... â1,q â∅,q ) -1 , a word of length p + q that is right-R R R-reversible to a word of length p + q + pq.

Other upper bounds can be obtained by using the action of F + sym on trees. To state the result, it is convenient to introduce the following natural terminology. Definition 10. For every signed A A A-word w, the right-numerator N r (w) and the rightdenominator D r (w) of w are the unique A A A-words satisfying w R R R N r (w)D r (w) -1 . Symmetrically, the left-numerator N ℓ (w) and the left-denominator D r (w) of w are the unique

A A A-words satisfying w R R R D ℓ (w) -1 N ℓ (w).
As left-and right-R R R-reversings are terminating, the positive A A A-words N r (w), D r (w), N ℓ (w), and D ℓ (w) exist for every signed A A A-word w. Proposition 3.15. Assume that w is a signed A A A-word and T * w is defined for some size n tree T . Then we have

(3.9) max(|N ℓ (w)| + |D r (w)|, |N ℓ (w)| + |D r (w)|) (n -1)(n -2)/2.
In order to establish Proposition 3.15, we need a preliminary result about the action of A A A-words on trees. First, if T is a tree and w is a signed A A A-word, we say that T * w is defined if T * u is defined for every prefix u of w. Now, if two signed A A A-words w, w ′ represent the same element of F , the hypothesis that T * w is defined for some tree T does not guarantee that T * w ′ is also defined: for instance, T * ε is always defined, but T * a -1 α a α is not. However, this cannot happen with reversing. Lemma 3.16. Assume that w, w ′ are signed-A A A-words and w is right-or left-R R R-reversible to w ′ . Then, for every tree T such that T * w is defined, T * w ′ is defined as well.

Proof. The problem with arbitrary equivalences is that new pairs a -1 α a α or a α a -1 α may be created. This however is impossible in the case of (right-or left-) reversing, as we can only delete such pairs, but not create them. A complete formal proof requires to check all possible cases: this is easy, and we skip the details.

∅ ≡ + R R R uN ℓ (w)D r (w)v ≡ + R R R uD ℓ (w)N r (w)v.
Then the function λ of (3.3) provides an upper bound for the lengths of the words R R Requivalent to a given word. In the current case, we have λ(a n-2 ∅ ) = (n -1)(n -2)/2, and (3.9) follows.

T T ′ C n C n w R R R R R R N ℓ (w) D r (w) N r (w) D ℓ (w) u v T ℓ T r Figure 16
. Bounding the lengths of the left-and right-numerators and denom- inators of a signed A A A-word w in terms of the size of a term T such that T * w is defined.

The upper bound of (3.9) is close to sharp: for w = (a 1 p-1 ... a 1 a ∅ ) -1 a q 1 p , the word D r (w) is â1 p-1 ,q ... â1,q â∅,q , which has length pq in the alphabet A A A, so the sum of the lengths of N ℓ (w) and D r (w) is p + pq, while the minimal size of a term T such that T * w is defined is p + q + 2.

To conclude with subword reversing, we mention one more result that involves both left-and right-reversing. The example of ε and a ∅ a -1 ∅ shows that R R R-equivalent words need not have R R R-equivalent numerators and denominators: the right-numerator of ε is ε, whereas the right-numerator of a ∅ a -1 ∅ is a ∅ . This cannot happen when left-and rightnumerators are mixed in a double reversing. Proposition 3.17. For w a signed A A A-word, define N ℓr (w) to be N r (D ℓ (w) -1 N ℓ (w)) and

D ℓr (w) to be D r (D ℓ (w) -1 N ℓ (w)). Then w ≡ R R R w ′ implies N ℓr (w ′ ) ≡ + R R R N ℓr (w) and D ℓr (w ′ ) ≡ + R R R D ℓr (w).
We first observe that N ℓr (w)D ℓr (w) -1 is a minimal fractionary expression of w: Lemma 3.18. If w, w ′ are R R R-equivalent signed A A A-words, there exist a positive A A A-word u satisfying

(3.11) N r (w ′ ) ≡ + R R R N ℓr (w) u and D r (w ′ ) ≡ + R R R D ℓr (w) u. Proof.
By construction, the word w is R R R-equivalent to D ℓ (w) -1 N ℓ (w), and the latter word is right-R R R-reversible to N ℓr (w)D ℓr (w) -1 . Hence we have (3.12) D ℓ (w)N ℓr (w) ≡ + R R R N ℓ (w)D ℓr (w), and, moreover, as mentioned in the proof of Proposition 3.11, the element of F + sym represented by D ℓ (w)N ℓr (w) and N ℓ (w)D ℓr (w) is the right-lcm of N ℓ (w) and D ℓ (w).

On the other hand, N r (w

′ )D r (w ′ ) -1 is R R R-equivalent to w ′ , hence to w, and therefore to D ℓ (w) -1 N ℓ (w) -1 . We deduce D ℓ (w)N r (w ′ ) ≡ R R R N ℓ (w)D r (w ′ ), whence (3.13) D ℓ (w)N r (w ′ ) ≡ + R R R N ℓ (w)D r (w ′ ) since F +
sym embeds in F . As D ℓ (w)N ℓr (w) is the right-lcm of N ℓ (w) and D ℓ (w), comparing (3.12) and (3.13) implies the existence of u satisfying (3.11).

of Proposition 3.17. By Lemma 3.18, there exist positive A A A-words u and u ′ satisfying

N ℓr (w ′ ) ≡ + R R R N ℓr (w)u and N ℓr (w) ≡ + R R R N ℓr (w ′ )u ′ , whence N ℓr (w) ≡ + R R R N ℓr (w)uu ′ . As F + sym is left-cancellative, we deduce ε ≡ + R R R uu ′ .
The only possibility is then that u and u ′ are empty.

We now return to Tamari lattices, and show how to use right-reversing to compute lowest upper bounds in the Tamari poset T n appealing to the words c T of (3.8). Of course, left-reversing can be used symmetrically to compute greatest lower bounds. Proposition 3.19. Assume that T, T ′ are size n trees. Then the least upper bound T ′′ of T and T ′ in the Tamari lattice T n is determined by

T ′′ = T * N r (c -1 T c T ′ ) = T ′ * D r (c -1 T c T ′ ).
Proof. As mentioned in the proof of Proposition 3.11, the words c T N r (c -1 T c T ′ ) and c T ′ D r (c -1 T c T ′ ) both represent the right-lcm of c T and c T ′ in F + sym. Hence, owing to Proposition 3.9, the image of c T N r (c -1 T c T ′ ) under I n , which, by definition, is

C n * c T N r (c -1 T c T ′ ), that is, T * N r (c -1 T c T ′ ), is the least upper bound in T n of I n (c T ), that is, of C n * c T , which is T , and I n (c T ′ ), that is, of C n * c T ′ , which is T ′ . Example 3.20. Let T = •(((•(••))•)•) and T ′ = (•(••))(•(••)). Using the method explained in Lemma 4.2 below, one obtains c T = a 11 a 2 1 and c T ′ = a 1 a ∅ . Right-reversing a -2 1 a -1
11 a 1 a ∅ leads to a 100 a ∅ a 0 a 00 a -2 ∅ (see Figure 17), and we deduce that the least upper bound of T and T ′ in the Tamari poset is the tree T * a 100 a ∅ a 0 a 00 , namely ((( T by right-reversing determines the least upper bound of T and T ′ in the Tamari lattice, here for

•(••))•)•)• (which is also T ′ * a 2 ∅ ).
T = •(((•(••))•)•) and T ′ = (•(••))(•(••)).

The Polish normal form on F

We now develop another approach for determining least common multiples in the monoid F + sym, whence, equivalently, least upper bounds in the Tamari lattices, namely using what is known as the Polish algorithm. Initially introduced in the case of the self-distributivity law [8, Chapter IX], the latter is easily adapted to our current context, where it provides a unique normal form for the elements of F + sym and a method for determining the upper bound of two trees in the Tamari lattice. The main technical tool here is the covering relation of [START_REF]On the rotation distance between binary trees[END_REF], a variant of the weight sequences of [START_REF]An algorithm to compute the Möbius function of the rotation lattice of binary trees[END_REF]-also see [START_REF] Pallo | Enumerating, ranking and unranking binary trees[END_REF] and [START_REF] Baril | Efficient lower and upper bounds of the diagonal-flip distance between triangulations[END_REF].

The section is organized as follows. In Subsection 4.1 we recall the standard Polish encoding of trees and its connection with the Tamari ordering. In Subsection 4.2 we describe an algorithm that, starting with the Polish encoding of two trees, determines a common upper bound of the latter in the Tamari lattice together with a distinguished way of performing the rotations. Then, in Subsection 4.3, we use the covering relation to control the previous algorithm and, in particular, prove that it always determines the least upper bound of the initial trees. Finally, in Subsection 4.4, we deduce a unique normal form for the elements of F that enjoys a sort of weak rationality property. 4.1. The Polish encoding of trees. As is well known, trees or, equivalently, parenthesized expressions can be encoded without parentheses using the Polish notation. Here we consider the right version, and use • as the operation symbol. Definition 11. For T a tree, the (right)-Polish encoding of T is the word T recursively defined by T = T if T has size 0, and

T = T 0 T 1 • for T = T 0 ∧ T 1 .
For T of size n, the Polish encoding T is a word of length 2n + 1, which we consider as a map of {1, ..., 2n + 1} into {•, •} (when we restrict to unlabeled trees): thus T (k) refers to the kth letter of the word T . There exists a natural one-to-one origin function from the positions of the letters of T to the addresses of the nodes of T , recursively defined for T = T 0 ∧ T 1 with T 0 of size n 0 by the rule that the origin of k in T is 0α where α is the origin of k in T 0 for k 2n 0 + 1, it is 1α where α is the origin of k -2n 0 -1 in T 1 for 2n 0 + 1 < k 2n, and it is ∅ for k = 2n + 1. For instance, the Polish encoding of the tree

•((••)•) of Figure 2 is •••••••,
and the corresponding origins are

• (0) • (100) • (101) • (10) • (11) • (1) • (∅) .
For our current purpose, it is important to note the following connection between the Polish encoding and the Tamari order. Proof. When translated to the right Polish notation, applying a left-rotation in a tree amounts to replacing some subword of the form T 0 T 1 T 2 •• with the corresponding word T 0 T 1 • T 2 •. The latter word is < Lex -larger than the former, as the beginning of the word is preserved, until the first letter • associated with T 2 , which is replaced with •.

When the initial letter • is erased, the words that are the Polish encoding of a trees identify with Dyck words, defined as those words in the alphabet {•, •} such that no initial segment has more •'s than •'s, see for instance [START_REF] Stanley | Enumerative Combinatorics[END_REF]. Using the standard correspondence between such words and random walks in N 2 , we obtain a simple receipe for determining the elements c T and c ′ T of (3.8) from T . Lemma 4.2. (See Figure 18.) Assume that T is a size n tree. For k in {1, ..., 2n + 1} recursively define ν T (k) by ν T (1) = -1 and, for k 2, (4.1)

ν T (k) =      ν T (k -1) + 1 for T (k -1) = T (k) = •, ν T (k -1) -1 for T (p -1) = T (p) = •, ν T (k -1) otherwise.
Then c ′ T is obtained from T by replacing each letter T (k) with ε if it is • and with a 1 i with i = ν T (k) if is it •; the word c T is obtained similarly after erasing the last block of •.

We skip the verification, a comparison of the recursive definitions of T and c ′ T . 4.2. The Polish algorithm. Assume that T, T ′ are trees of size n and we look for a (minimal) tree T ′′ that is an upper bound of T and T ′ in the Tamari order. If T and T ′ do not coincide, then one of the words T , T ′ is lexicographically smaller than the other, say for instance T . This means means that there exists k such that T (k) is •, whereas T ′ (k) is •. In this case, we shall say that T and T ′ have a clash at k. Here is the point. Lemma 4.3. Assume that T is a tree and that the kth letter in T is •. Then there exists at most one pair (α, r) such that T * âα,p is defined and T and T * âα,r have a clash at k. Moreover, if there exists T ′′ such that T and T ′′ have a clash at k, there exists exactly one pair as above.

-1 0 1 2 • • • • • • • • • • • a ∅ a 11 a 1 a 1 a ∅
Proof. As Figure 19 shows, if we have T ′ = T * âα,r , then the words T and T ′ coincide up to the first letter coming from the α10 r-1 1-subtree of T : the latter is • (as is always the first letter of a Polish encoding), whereas, in T ′ , we have a letter • at this position. Thus, the action of âα,r on T is to replace • by • at a position whose origin in T has the form α10 r-1 10 i for some i.

Consider now the kth letter in T , supposed to be a letter •. The origin of k in T is a certain address of leaf in T , say β. By the above argument, a pair (α, r) may result in a clash at k only if we can write β = α10 r-1 10 i for some r 1 and i 0. For every β, this happens for at most one pair (α, r), and this happens if and only if β contains at least two digits 1.

Assume now that T and T ′′ have a clash at k, and consider the value of ν T (k) as defined in (4.1). By construction (and by the standard properties of Dyck words), we have ν T ′′ (k) 0 as T ′′ (k) is •. By construction, we have ν T (k) > ν T ′′ (k) since T (k) is •, whence ν T (k) 1. This implies (actually an equivalence) that the address β contains at least two digits 1. Hence there exists a pair (α, r) as above. In particular, if T, T ′ are trees satisfying T T T ′ , it is not a priori clear that the Polish algorithm terminates with the pair (T ′ , T ′ ), that is, the clashes always occur on the first of the two current trees. We shall see now that this is actually true. The main tool will be the covering relation, a binary relation that provides a description of the shape of a tree in terms of the addresses of its leaves. We recall that, if T is a size n tree, T # denotes the labeled tree obtained by attributing to the leaves of T labels 0 to n from left to right. So, for instance, for

T T ′ T 0 T 1 T 2 T r T r+1 âα,r α α α1 α10 α10 r α10 r-1 α10 r-1 1 T 0 T 1 T 2 T r T r+1 ↓ ↓ ↓ ↓ ↓ ↓ ↑ j0 ↑ j1 ↑ j2 ↑ jr ↑ j0 ↑ j1 ↑ j2 ↑ jr
T = •((••)•), we have T # = • 0 ((• 1 • 2 )• 3 ), and T # = • 0 • 1 • 2 •• 3 ••.
Definition 13. (See Figure 20.) Assume that T is a size n tree. For 0 i n, we define add T (i) to be the origin of • i in T # . Then, for j > i, we say that j covers i in T , written j ⊲ T i, if there exists an address γ such that add T (j) has the form γ1 p for some positive p and add T (i) begins with γ0. We write j T i for "j ⊲ T i or j = i". covers i in T if there exists a subtree T ′ such that j is the last (rightmost) label in T ′ , whereas i is a non-final label in T ′ . For instance, in the right hand tree, 4 covers 1, 2, 3, but does not cover 0, and 3 covers 1 and 2, whereas 2 covers nobody.

It is easily seen [START_REF]On the rotation distance between binary trees[END_REF] that, for every j occurring in a tree T , the set of all i's covered by j is either empty or is an interval ending in j -1: if j ⊲ T i and j > i ′ i hold, then so does j ⊲ T i ′ . Also, the relation ⊲ T is transitive, and it determines T . We shall need the more precise result that every initial fragment of the covering relation determines the corresponding initial fragment of the Polish encoding of T . Lemma 4.6. Assume that T is a size n tree. Then, for 1 j n + 1, the number of symbols • following the jth letter • in T is the number of i's satisfying j ⊲ T i and k ⊲ T i for j > k > i.

Proof. Write j ⊲ #

T i if we have j ⊲ T i and k ⊲ T i for j > k > i. Then, by definition, j ⊲ # T i holds if and only we have add T (j) = α1 q and add T (i) = α01 p for some α and some p, q 0. Indeed, if we have add T (i) = α0β with β containing at least one 0, say β = 1 p 0γ, then we have k ⊲ T i for k satisfying add T (k) = α01 p 01 r .

On the other hand, an induction shows that the jth letter • in T is followed by r letters • if and only if the address add T (j) has the form γ1 r for some γ that does not finish with 1, that is, is empty or finishes with 0. Now, assume that add T (j) is γ1 r . For 0 m < r, let i m be the (unique) position whose address in T has the form γ1 m 01 q for some q. By the above characterization, we have j ⊲ # T i m . So the number of i's satisfying j ⊲ # T i is at least r.

Lemma 4.10. Assume that T, T ′ are equal size trees. Then the Polish algorithm running on (T, T ′ ) terminates with a pair (T ∞ , T ∞ ) such that ⊲ T∞ is the transitive closure of ⊲ T and ⊲ T ′ .

Proof. Let (T t , T ′ t ) denote the pair of trees obtained after t steps of the Polish algorithm running on (T, T ′ ), and N be the total number of steps. By Lemma 4.8, the relations ⊲ Tt make a non-decreasing sequence with respect to inclusion, and so do the relations ⊲ T ′ t . So, in particular, the transitive closure of ⊲ T and ⊲ T ′ is included in the transitive closure of ⊲ T N and ⊲ T ′ N . Now, by hypothesis, the latter is ⊲ T∞ . On the other hand, Lemma 4.9 shows that, for every t, the relation ⊲ Tt+1 is included in the transitive closure of ⊲ Tt and ⊲ T ′ t , and so is ⊲ T ′ t+1 . Hence the transitive closure of ⊲ Tt+1 and ⊲ T ′ t+1 is the transitive closure of ⊲ Tt and ⊲ T ′ t . Hence ⊲ T∞ , which is the transitive closure of ⊲ T N and ⊲ T ′ N , is the transitive closure of ⊲ T0 and ⊲ T ′ 0 .

We are ready to put pieces together and state the main results of this section.

Proposition 4.11. For T, T ′ , T ′′ equal size trees, the following are equivalent: (i) The tree T ′′ is the least upper bound of T and T ′ in the Tamari lattice;

(ii) The Polish algorithm running on (T, T ′ ) returns (T ′′ , T ′′ );

(iii) The covering relation of T ′′ is the transitive closure of those of T and T ′ .

Proof. Let T ∨ be the least upper bound of T and T ′ in the Tamari lattice, and T ∞ be the tree such that the Polish algorithm running on (T, T ′ ) returns (T ∞ , T ∞ ). By Lemma 4.8, ⊲ T and ⊲ T ′ are included in ⊲ T∨ . Hence the transitive closure of ⊲ T and ⊲ T ′ , which by Lemma 4.10 is ⊲ T∞ , is included in ⊲ T∨ .

On the other hand, by definition, we have T T T ∞ and T ′ T T ∞ , whence T ∨ T T ∞ . This implies that ⊲ T∨ is included in ⊲ T∞ . Hence ⊲ T∨ and ⊲ T∞ coincide, and, therefore, T ∨ = T ∞ holds. So (i) and (ii) are equivalent.

Next, as said above, (ii) implies (iii) by Lemma 4.8. Conversely, if T ′′ is such that ⊲ T ′′ is the transitive closure of ⊲ T and ⊲ T ′ , then ⊲ T ′′ coincides with ⊲ T∞ and, therefore, we must have T ′′ = T ∞ . So, (ii) and (iii) are equivalent.

Corollary 4.12. For T, T ′ equal size trees, the following are equivalent:

(i) We have T T T ′ in the Tamari order;

(ii) There exists f in F + sym such that T ′ = T * f holds; (iii) S(T, T ′ ) is a positive A A A-word, that is, the Polish algorithm running on (T, T ′ ) finishes with (T ′ , T ′ ).

(iv) The relation

⊲ T is included in ⊲ T ′ .
Proof. The equivalence of (i) and (ii) has been established in Lemma 3.1. Next, it is obvious that (iii) implies (ii) as every element âα,r belongs to F + sym. Conversely, if T T T ′ holds, then the least upper bound of T and T ′ is T ′ . Hence, by Proposition 4.11, the Polish algorithm running on (T, T ′ ) finishes with (T ′ , T ′ ). This means that the word S(T, T ′ ) contains positive letters âα,r only. So (ii) implies (iii).

Finally, as observed above, (iii) is equivalent to saying that the Polish algorithm running on (T, T ′ ) finishes with (T ′ , T ′ ), whereas (iv) is equivalent to saying that ⊲ T ′ is the transitive closure of ⊲ T and ⊲ T ′ . By Proposition 4.11, the latter properties are equivalent and, therefore, (iii) and (iv) are equivalent.

It should be noted that the equivalence of (i) and (iv) in Corollary 4.12 already appears as [23, Theorem 2.1].

4.4.

The Polish normal form. One of the interests of Proposition 4.11 and Corollary 4.12 is that they provide unique distinguished decompositions for every element of F and of F + sym in terms of the generators âα,r . Indeed, we obtained for every pair of equal size trees (T, T ′ ) a certain signed A A A-word S(T, T ′ ) such that T * S(T, T ′ ) is defined and equal to T ′ . This word S(T, T ′ ) does not depend on T . Lemma 4.13. Assume that f belongs to F and T * f is defined. Then the signed A A Aword S(T, T * f ) is an expression of f , and it does not depend on T .

Proof. First, we have T * f = T * S(T, T ′ ), so, by Proposition 2.4, the word S(T, T ′ ) is an expression of f . Next, assume that σ is a substitution, and let us compare the Polish algorithm running on a pair (T, T ′ ) and on the pair (T σ , T ′σ ). The word T σ is obtained from the word T by replacing every variable • i with the corresponding word σ(i) . As the variables occur in the same order in the words T and T ′ , substituting • i with σ(i) introduces no new clash. Therefore, if (T t , T ′ t ) are the trees at the tth step of the algorithm running on (T, T ′ ), then (T σ t , T ′ t σ ) are the trees at the tth step of the algorithm running on (T σ , T ′σ ), implying S(T, T ′ ) = S(T σ , T ′σ ).

By definition, for every f in F , there exists a unique pair of trees (f -, f + ) such that every pair (T, T * f ) can be expressed as ((f # -) σ , (f # + ) σ ). The above result then shows that S(T, T * f ) coincides with S(f -, f + ), which only depends on f . Definition 14. For f in F , the Polish normal form of f is the signed

A A A-word S(f -, f + ). Example 4.14. Let f = a ∅ a 1 = a 11 a ∅ (= x 1 x 2 = x 3 x 1 ). Then we have f -= •(•(•(••))) and f + = (••)((••)•).
Running the Polish algorithm on these trees returns the (positive) A A A-word a ∅ a 1 : so the latter is the Polish normal form of f . By contrast, a 11 a ∅ , which is another A A A-expression of f , is not normal. One verifies similarly that the word a 2 ∅ is normal, whereas the equivalent words a 1 â∅,2 and a 1 a ∅ a 0 are not. Corollary 4.12 immediately implies: Proposition 4.15. An element of F belongs to the submonoid F + sym if and only if its Polish normal form contains no letter â-1 α,r .

As the family of generators A A A is infinite, it makes no sense to wonder whether Polish normal words form a rational language or whether the Polish normal form can be connected with an automatic structure. However, let us observe that being Polish normal is a local property that can be characterized in terms of adjacent letters. Proposition 4.16. A positive A A A-word âα1,r1 ... âα ℓ ,r ℓ is Polish normal if and only if α t 0 rt < α t+1 10 rt+1-1 1 holds for every t < ℓ, where < denotes the left-right (partial) ordering of addresses.

Proof. Let w = âα1,r1 ... âα ℓ ,r ℓ and assume that T * w is defined. For 0 t ℓ, put T t = T * âα1,r1 ... âαt,rt . Then w is normal if, for every 1 t ℓ, we have âαt,rt = s(T t-1 , T ℓ ), that is, âαt,rt appears at the tth step of the Polish algorithm running on (T, T * w). Now, as shown in Figure 19, the origin in T t of the letter • involved in the clash between T t-1 and T t is α t 0 rt , whereas the origin in T t of the letter • involved in the clash between T t and T t+1 lies in α t+1 10 rt+1-1 1{0} * . The normality condition is then that, in T t , the former letter lies on the left of the latter. By construction of the Polish encoding, this happens if and only if the first address precedes the second in the "left-right-root" linear ordering of addresses. Due to the form of the second address, this is equivalent to α t 0 rt < α t+1 10 rt+1-1 1.

Syntactic invariants.

A natural way to investigate distances in the Tamari lattices is to use the action of F on trees and to study the length of the elements of F with respect to the generating family A A A. Indeed, Proposition 2.4 directly implies Lemma 5.4. For all trees T, T ′ , we have dist(T, T ′ ) = S(T, T ′ ) A A A , where f A A A is the A A A-length of f , that is, the length of the shortest signed A A A-word representing f .

In order to establish (lower) bounds on f A A A , a natural approach is to use the syntatic properties of the relations of R R R. 11 a α a -1 11 . Left-reversing the word a 1 a -1 11 a α a -1 11 yields the word a -1 111 a -1 1111 a 1 a ∅ , which in turn is right-reversible to a 1 a ∅ a -1 1 a -1 11 . We conclude that N ℓr (w) is a 1 a ∅ and D ℓr (w) is a 11 a 1 . Then Proposition 5.6 gives f A A A 4, that is, the word a 1 a -1 11 a α a -1 11 is geodesic. The numerator is the word up of Lemma 5.9, whereas the denominator is the length p word â∅,p (as usual, the letters "a" have been skipped).

at least one step with critical index 2p + 1, the critical index of a α being defined as the unique j such that applying a α adds at least one relation j ⊲ i, that is, the unique j such that add T (j) lies in α10{1} * . Now, here is the point: 2p covers 0 and 1 in T ′ p , but not in T p . We deduce that every A A A-word transforming T p to T ′ p contains at least one step with critical index 2p. But we claim that every such word must actually contain at least two such steps, that is, one step cannot be responsible for the two new covering relations. Indeed, 2p covers 2 in T p , but does not cover 1. The only situation when a step adding 2p ⊲ 1 in a tree T can simultaneously add 2p ⊲ 0 is when 1 covers 0 in T . But this cannot happen here, because we are considering positive A A A-words only, so any possible covering satisfied at an intermediate step must remain in the final tree T ′ p . As 1 does not cover 0 in T ′ p , it is impossible that 1 covers 0 at any intermediate step. Thus two steps are needed to ensure 2p ⊲ 1 and 2p ⊲ 0.

The sequel is similar. In T ′ p , the number 2p -1 covers 0, 1, and 2, whereas, in T p , it covers only 3. Then at least three steps are needed to ensure 2p -1 ⊲ 2, 2p -1 ⊲ 1, and 2p -1 ⊲ 0. Indeed, a step can cause 2p -1 to simultaneously cover 1 and 2 only if 2 covers 1 at the involved step, which cannot happen as 2 does not cover 1 in T ′ p ; the argument is then the same for 1 and 0 once we know that 2 does not cover 1.

Similarly, 4 steps are needed to force 2p -2 to cover 3 to 0, then 5 steps to force 2p -3 to cover 4 to 0, etc., and p + 1 steps to force p + 1 to cover p to 0. We conclude that 1 + 2 + • • • + (p + 1), that is, (p + 1)(p + 2)/2, steps at least are needed to go from T p to T ′ p . Hence u p is geodesic among positive A A A-words. Thus the proof of Proposition 5.8 is complete. Finally, we translate the above arguments into the language of Tamari lattices. Definition 17. For T, T ′ in T n satisfying T T T ′ , the positive distance dist + (T, T ′ ) from T to T ′ is the minimal number of left-rotations needed to transform T into T ′ . Proposition 5.12. For every even n, there exist T, T ′ in T n satisfying (5.2) dist(T, T ′ ) 12 n dist + (T, T ′ ).

Proof. Write n = 2p+2 and let T p , T ′ p be the size n trees of the proof of Lemma 5.11. Then we have dist(T p , T ′ p ) 3p + 1 and dist + (T p , T ′ p ) = (p + 1)(p + 2)/2, whence dist(T p , T ′ p )

12n-16 n(n+2) dist + (T p , T ′ p ) in term of n, and, a fortiori, (5.2). Corollary 5.13. Chains are not geodesic in Tamari lattices; more precisely, for every n, there exists a length ℓ chain of T n whose endpoints are at distance less than 12ℓ/n.

We conclude with a few open questions.

Question 5.14. For w a positive A A A-word, is the number λ(w) of (3.3) a least upper bound for the length of the words that are R R R-equivalent to w?

The answer is positive in the case of a n ∅ . Indeed, we have λ(a n ∅ ) = n(n + 1)/2 and a n ∅ ≡ + R R R â1 n-1 ,1 â1 n-2 ,2 ... â1,n-1 â∅,n . The general case is not known. Question 5.15. Does the Polish normal form of Definition 14 satisfy some Fellow Traveler Property, that is, is the distance between the paths associated with the normal forms of elements f and f a ±1 α uniformly bounded? A positive solution would provide a sort of infinitary automatic structure on F . The question should be connected with a possible closure of Polish normal words under leftor right-R R R-reversing Finally, using mapping class groups and cell decompositions, D. Krammer constructed for every size n tree T an exotic lattice structure on T n in which T is the bottom element [START_REF] Krammer | A class of Garside groupoid structures on the pure braid group[END_REF]. Question 5.16. Can the Krammer lattices be associated with submonoids of the Thompson group F ? More generally, a natural combinatorial description of the Krammer lattices is still missing, but would be highly desirable. Connections with permutations and braids in the line of [START_REF] Björner | Shellable nonpure complexes and posets. II[END_REF] can be expected.

Figure 1 .

 1 Figure 1. The Tamari lattice T3 made by the five ways of bracketing a four variable parenthesized expression.

Figure 4 .

 4 Figure 4. The Tamari lattice T4, both in terms of parenthesized expressions and binary trees.

Figure 5 .

 5 Figure 5. Two representations of the elements x0 and x1 of the Thompson group F : above, the usual graph of a function of [0, 1] into itself, below, a diagram displaying the two involved dyadic decompositions of [0, 1], with the source above and the target below: this simplified diagram specifies the function entirely.

x 0 x 1 Figure 6 .

 16 Figure 6. Canonical pair of trees associated with an element of the group F .

Figure 7 .

 7 Figure 7. Action of x0 and x1 on a tree: respectively applying a left-rotation at the root, and at the right-child of the root.

Figure 9 .Figure 10 .

 910 Figure 9. Quasi-commutation relation in F : the general scheme and one example.

Figure 13 .

 13 Figure 13. Closure of the family A A A under right-reversing: â-1 10,3 â∅,4 reverses to â∅,7â-1 0,3 , which corresponds to the last relation in (3.6) with α = ∅, r = 4, s = 3, i = 1 (the letter "a" has been skipped everywhere).

Figure 14 .

 14 Figure 14. Extended pentagon relation âα,r âα0 i ,s = âα10 i ,s âα,r+s, here for α = ∅, r = 4, s = 3, i = 2.

Figure 15

 15 Figure 15. For T a size n tree, cT describes how to construct T from the right-comb Cn, and c ′ T describes how to construct T ∧ Cp from Cn+p; the figure illustrates the recursive definition of c ′ T (above) and cT (below) for T = T0 ∧ T1, with n1 denoting the size of T1.

Proposition 3 . 11 .

 311 The poset (F + sym,

of Proposition 3 . 15 .

 315 Let T ′ = T * w. By definition, w is right-R R R-reversible to N r (w)D r (w) -1 , and left-R R R-reversible to D ℓ (w) -1 N ℓ (w). By Lemma 3.16, this implies that T * N r (w)D r (w) -1 and T * D ℓ (w) -1 N ℓ (w) are defined. Put T ℓ = T * D ℓ (w) -1 and T r = T * N r (w). By hypothesis, the terms T, T ′ , T ℓ , and T r all have size n. Hence there exists a positive A A A-word u (namely c T ℓ ) mapping the right comb C n to T ℓ . By symmetry, there exists a positive A A A-word v mapping T r to the left comb C n . Then uN ℓ (w)D r (w)v and uD ℓ (w)N r (w)v are R R R-equivalent positive A A A-words, and both map C n to C n , see Figure 16. Now a n-2 ∅ also maps C n to C n . Hence, by Proposition 2.4, we must have (3.10) a n-2

Figure 17 .

 17 Figure 17. Computing the right-lcm of cT and c ′

Lemma 4 . 1 .

 41 Let < Lex denote the lexicographical extension of the ordering • < • to {•, •}-words. Then, for all trees T, T ′ , the relation T T T ′ implies T Lex T ′ .

Figure 18 .

 18 Figure 18. Computing cT and c ′ T from the Polish encoding T of T : write the kth letter of T at level νT (k); then c ′ T is read from the levels of the letters •. Here, for (••)((•(••))•), we read c ′ T = a ∅ a11a 2 1 a ∅ , and, discarding the last two symbols •, cT = a ∅ a11a1.

Figure 19 .

 19 Figure[START_REF] Krammer | A class of Garside groupoid structures on the pure braid group[END_REF]. Action of âα,r: the Polish encodings coincide up to the first • corresponding to T2 in T (black arrow); the latter is replaced with • in T ′ because, in T ′ , there is one more right-edge after T1 than in T , and the clash occurs between the marked letters.

4. 3 .

 3 The covering relation. For the moment, we have no connection between the common upper bound of two trees provided by the Polish algorithm of Subsection 4.2 and their least upper bound in the Tamari lattice.

Figure 20 .

 20 Figure 20. Covering relation of T : the leaves are numbered 0 to n, and j

Figure 21 .

 21 Figure 21. The zigzag trees of Conjecture 5.3, here for n = 15; the distance is 24, as predicted.

Lemma 5 . 5 .Example 5 . 7 .

 5557 For w a signed A A A-word, denote by |w| 1 the number of letters a ±1 1 i in w.(i) If u, u ′ are R R R-equivalent positive A A A-words, then |u| 1 = |u ′ | 1 holds. (ii) If w, w ′ are signed A A A-words, then w R R R w ′ implies |w| 1 |w ′ | 1 holds.Proof. In both cases, it suffices to inspect the relations of R R R. In the case of the pentagon relations, we have |a 2 α | 1 = |a α0 a α a α1 | 1 , both being 2 for α in {1} * , and 0 otherwise. Similarly, for (ii), we find |a -1α a α1 | 1 = |a α a -1 α0 a -1 α | 1 , both being 2 for α in {1} * , and 0 otherwise. The inequality comes from |a -1 α a α | 1 = 2 > 0 = |ε| 1 for α in {1} * .Note that the counterpart of Lemma 5.5(ii) involving left-reversing is false:a ∅ a -1 0 is left-R R R-reversible to a -1∅ a 1 a ∅ , and we have |a ∅ a -10 | 1 = 1 < 3 = |a -1∅ a 1 a ∅ | 1 . Proposition 5.6. For every f of F and every A A A-word w representing f , we have(5.1) f A A A |D ℓr (w)| 1 + |N ℓr (w)| 1 . Proof. Put ℓ = |D ℓr (w)| 1 + |N ℓr (w)| 1 .By definition, the word w is right-R R R-reversible to the word N r (w)D r (w) -1 , so, by Lemma 5.5(ii), we have|w| 1 |N r (w)D r (w) -1 | 1 = |N r (w)| 1 + |D r (w)| 1 .Next, it follows from Proposition 3.17 that there exist a positive A A A-word u satisfying N r (w)≡ + R R R N ℓr (w) u and D r (w) ≡ + R R R D ℓr (w) u.Then, by Lemma 5.5(i), we deduce |N r (w)| 1 + |D r (w)| 1 ℓ, whence |w| |w| 1 ℓ. Now assume w ′ ≡ R R R w. By the result above, we have |w ′ | |N ℓr (w ′ )| 1 + |D ℓr (w ′ )| 1 . By Proposition 3.17, we have N ℓr (w ′ ) ≡ + R R R N ℓr (w) and D ℓr (w ′ ) ≡ + R R R D ℓr (w), whence, by Lemma 5.5(i), |N ℓr (w ′ )| 1 + |D ℓr (w ′ )| 1 = ℓ. Thus |w ′ | ℓ holds for every word w ′ that represents w. By definition, this means that f A A A ℓ is true. Of course, a symmetric criterion involving {0} * instead of {1} * may be stated. Let f = a 1 a -1

Figure 22 .

 22 Figure 22. Right-reversing the signed A A A-word wp, here with p = 3. The

Figure 23 .

 23 Figure 23. The trees of Lemma 5.11 with the leaf labelling involved in the covering relation, here in the case p = 3. The tree T ′′ p is the intermediate tree Tp * wp obtained after 2p + 1 steps: applying â∅,p to T ′′ p leads to T ′ p in a total of 3p + 1 rotations.

Work partially supported by the ANR grant TheoGar ANR-08-BLAN-0269-02.

1

Now the principle of an algorithm should be clear: starting with two trees T, T ′ such that the Polish encoding T and T ′ coincide up to position k -1, we have found a unique way of applying an iterated left-rotation âα,r to one of the trees so that the clash is moved further to the right. By iterating the process, we obtain after finitely many steps two trees whose Polish encodings coincide, that is, we obtain a common upper bound for the initial trees T, T ′ . Definition 12. Assume that T, T ′ are trees of equal size.

(i) If T < Lex T ′ holds, we denote by s(T, T ′ ) the unique element âα,r such that T * âα,r and T ′ have no clash at the position where T and T ′ have one.

(ii) We denote by S(T, T ′ ) the signed A A A-word recursively defined by the rules . We find

0 , with a clash at 4 (underlined). The origin of 4 in T 0 is 10011, whence s(T 0 , T ′ 0 ) = a 100 , and

1 , with a clash at 5. The origin of 5 in T 1 is 1001, whence s(T 1 , T ′ 1 ) = â∅,3 , and S(T 1 , T ′ 1 ) = â∅,3 S(T 2 , T ′ 2 ) with T 2 = T 1 * â∅,3 and

with a clash at 7. The origin of 7 in

, with a clash at 9. The origin of 9 in T ′ 3 is 11, whence s(T ′ 3 , T 3 ) = a ∅ , and

4 , so the algorithm halts. The tree T 4 is a common upper bound of T 0 and T ′ 0 , and the word S(T 0 , T ′ 0 ) is a 100 â∅,3 a -2 ∅ .

Thus, for all equal size trees T, T ′ , we obtained a distinguished signed A A A-word S(T, T ′ ), and, by construction, the relation T ′ = T * S(T, T ′ ) is satisfied. Remark 4.5. As mentioned in the beginning of the section, an entirely similar algorithm can be defined with the self-distributivity law x(yz) = (xy)(xz) replacing the associativity law x(yz) = (xy)z. Then tree rotations are replaced with distributions, which consist in replacing subtrees T 0 ∧ (T 1 ∧ T 2 ) with (T 0 ∧ T 1 ) ∧ (T 0 ∧ T 2 ). In this case, the size of the trees is changed by the transformations, and termination becomes problematic. Actually, in spite of experimental evidence [START_REF] Deiser | Notes on the Polish Algorithm[END_REF] and positive partial results [START_REF]Braids and Self-Distributivity[END_REF], the question, which seems to be extremely difficult, remains open.

Conversely, assume that there are r different positions i 0 < ... < i r-1 satisfying j ⊲ # T i m . By the above characterization, there exist α 0 , ..., α r-1 satisfying add T (i m ) = α m 01 * and add T (j) = α1 * As the numbers i m are pairwise distinct, so are the addresses α m and, therefore, we have add T (j) = α 0 1 r ′ with r ′ r.

It directly follows from Lemma 4.6 that the covering relation of a tree T determines the Polish encoding of T , hence T itself. Actually, the lemma shows more. Lemma 4.7. Assume that T, T ′ are equal size trees, and (as a set of pairs) ⊲ T is properly included in ⊲ T ′ . Then T < Lex T ′ holds.

Proof. Let j be minimal such that there exists i satisfying j ⊲ T ′ i but not j ⊲ T i. For k < j, the restriction of the covering relations ⊲ T and ⊲ T ′ to the interval [1, k] coincide and, therefore, by Lemma 4.6, the numbers of symbols • following • k-1 in T # and T ′# are equal. So, up to • j-1 , the words T and T ′ coincide.

Consider now • j-1 . We claim that the number r ′ of • following • j-1 in T ′# is larger than its counterpart r in T # , resulting in a clash between T and T ′ and in the inequality T < Lex T ′ . To see that r ′ > r holds, we use Lemma 4.6 again. Using ⊲ # T as in the proof of Lemma 4.6, we note that j ⊲ # T i implies j ⊲ # T ′ i as the restrictions of ⊲ T and ⊲ T ′ to [1, j -1] coincide. By hypothesis, there are r values of i satisfying j ⊲ # T i, and these values also satisfy j ⊲ # T ′ i. Now, by hypothesis, there exists i ′ satisfying j

Hence there are strictly more than r values of i satisfying j ⊲ # T ′ i, as expected.

When a left-rotation transforms a tree T into a tree T ', the covering relation of T is included in that of T ′ . The precise relation is as follows. Hereafter we use {0, 1} * (resp. {1} * ) for the set of all addresses (resp. all addresses of the form 1 i ). Proof. Consider Figure 19 again. Let j 1 , ..., j r denote the last variable in the subtrees T 1 , ..., T r . A direct inspection shows that every covering pair in T is still a covering pair in T ′ , and that the new covering pairs are the pairs (j m , i) with 1 m r and i occurring in T 0 : the action of âα,r is to let j 1 , ..., j m cover the variables of T 0 . Converted into addresses, this gives (4.3).

Lemma 4.8 is important for the Polish algorithm as it bounds possible coverings. Proof. Assume s(T, T ′ ) = âα,r and let T 1 = T * âα,r . We use the notation of Figure 19 once more, calling j m the rightmost variable occurring in T m for 0 m r. Let I denote the set of all i's occurring in the subtree T 0 . By Lemma 4.8, the pairs that belong to ⊲ T1 and not to ⊲ T are the pairs (j 1 , i), ..., (j r , i) with i in I. The hypothesis that âα,r is s(T, T ′ ) implies that the number of • following • j-1 in T ′# is larger than its counterpart in T , so j 1 must cover strictly more positions in T ′ than in T . So, necessarily, j 1 ⊲ T ′ j 0 holds. On the other hand, j 0 T i holds for every i in I, and j m T j 1 holds for 1 m r. It follows that, for all m in {1, ..., r} and i in I, the pair (j m , i) belongs to the the transitive closure of ⊲ T and ⊲ T ′ .

For instance, the word a ∅ a ∅ is normal, as we have ∅0 1 = 0 < ∅10 1-1 1 = 11, but a 1 â∅,2 is not, as we do not have 10 1 = 10 < ∅10 2-1 1 = 101.

Distance in Tamari lattices

We conclude this description of the connections between the Tamari lattice and the Thompson group F with a few observations about distances in T n . The general principle is that it is easy to obtain upper bounds, but difficult to prove lower bounds and many questions remain open in this area. Our main observation here is that the embedding of the monoid F + sym into the group F is not an isometry, and not even a quasi-isometry (Definition 16): for every positive constant C, there exist elements of F + sym whose length in F is smaller than their length in F + sym by a factor at least C. In terms of Tamari lattices, this implies that chains are not geodesic (Corollary 5.13).

The plan of the section is as follows. In Subsection 5.1, we quickly survey the known results about the diameter of Tamari lattices. Then, we show in Subsection 5.2 how to use the syntactic relations of R R R to obtain (rather weak) distance lower bounds. Finally, in Subsection 5.3, we use the covering relation to establish (stronger) lower bounds.

5.1. The diameter of n . Surprisingly, the diameter of the Tamari lattice T n is not known for every n.

Definition 15. For T, T ′ in T n , the distance between T and T ′ , denoted by dist(T, T ′ ) is the minimal number of rotations needed to transform T into T ′ . The diameter of T n is the maximum of dist(T, T ′ ) for T, T ′ in T n .

Theorem 5.1 (Sleator, Tarjan, Thurston [START_REF] Sleator | Rotation distance, triangulations, and hyperbolic geometry[END_REF]). For n 11, the diameter of T n is at most 2n -6; for n large enough, it is exactly 2n -6.

The argument uses the fact that the maximal distance between two size n trees is also the maximal number of flips needed to transform two triangulations of an (n + 2)-gon one into the other. A lower bound for the latter is obtained by putting the considered triangulations on the two halves of a sphere and bounding the hyperbolic volume of the resulting tiled polyhedron. It is conjectured that the value 2n -6 is correct for every n 11. However, due to its geometric nature, the argument of [START_REF] Sleator | Rotation distance, triangulations, and hyperbolic geometry[END_REF] works only for n n 0 , with no estimation of n 0 .

By contrast, combinatorial arguments involving the covering relation of Subsection 4.3 lead to (weaker) results that are valid for every n. Theorem 5.2. [START_REF]On the rotation distance between binary trees[END_REF] For n = 2p 2 , the diameter of T n is at least 2n -2 √ 2n + 1 and, for every n, it is at least 2n -√ 70n.

Although no theoretical obstruction seems to exist, the covering arguments have not yet been developed enough to lead to an exact value of the diameter. However some candidates for realizing the maximal distance are known. Conjecture 5.3. [START_REF]On the rotation distance between binary trees[END_REF] For α an address, let α denote the tree recursively specified by the rules ∅ = •, 0α = α ∧ •, and 1α = • ∧ α . Define

for n = 2p + 3, 000(10) p-2 1 for n = 2p + 4, see Figure 21. Then one has dist(Z n , Z ′ n ) = 2n -6 for n 9. Conjecture 5.3 has been checked up to size 19 (sizes below 9 are special, because the trees are then too small for the generic scheme to start; by the way, the value 2n -6 is valid for n = 5, 6, 7, but not for n ≤ 4 and for n = 8).

By construction, the elements c T involved in proof of Proposition 3.9 are represented by A A A-words all letters of which are of the form a α with α in {1} * , and, therefore, these words are geodesic. However, elements of this type are quite special, and the criterion of Proposition 5.6 is rarely useful. In particular, it follows from the construction that every element of F can be represented by a word of the form c -1 T c T ′ but, even when the fraction is irreducible, that is, when the elements represented by c T and c T ′ admit no common left-divisor in F + sym, it need not be geodesic, as shows the example of a -p ∅ a 1 p a 1 p-1 ... a 1 , an irreducible fraction of length 2p which is R R R-equivalent to the positive-negative word a ∅ a -p 0 a -1 ∅ of length p + 2. 5.3. The embedding of F + sym into F . More powerful results can be obtained using the covering relation of Subsection 4.3. As an example, we shall now establish that the embedding of the monoid F + sym into the group F provided by Proposition 3.2 is not an isometry, that is, there exist elements of F + sym whose length as elements of F is smaller than their length as elements of F + sym. This result is slightly surprising: clearly, fractions need not be geodesic in general, but we might expect that, when an element of F belongs to F + sym, then its length inside F + sym equals its length inside F . Definition 16. If (X, d), (X ′ , d ′ ) are metric spaces, a map f : X → X ′ is a quasi-isometry if there exist C 1 and

The result we shall prove is as follows.

Proposition 5.8. For f in F + sym, let f + A A A denotes the A A A-length of f in F + sym, that is, the length of a shortest positive A A A-word representing f . Then the embedding of F + sym into F is not a quasi-isometry of (F + sym,

In order to establish Proposition 5.8, it is enough to exhibit a sequence of elements

This is what the next result provides.

Lemma 5.9. For every p 1, let u p be the A A A-word â(10) p ,1 â(10) p-1 ,2 ... â10,p â∅,p+1 .

Then, for every p, we have u p A A A 3p + 1 and u p + A A A = (p + 1)(p + 2)/2. Establishing Lemma 5.9 requires to prove two inequalities, namely an upper bound on u p A A A and a lower bound on u p + A A A . As always, the first task is easier than the second. Lemma 5.10. For every p 1, we have u p A A A 3p + 1.

Proof. For p 1, let w p = a (10) p a -1 (10) p-1 1 a (10) p-1 a -1 (10) p-2 1 a (10) p-2 ... a -1 101 a 10 a -1 1 a ∅ . Then w p is a signed A A A-word of length 2p + 1. An easy induction using the formulas of (3.6) shows that w p is right-reversible to the positive-negative word u p â-1 ∅,p , see Figure 22. As the latter word has A A A-length 3p + 1, the result follows.

In order to complete the proof of Lemma 5.9, it remains to prove that u p + A A A is p(p + 3)/2. As the length of u p is p(p + 3)/2, the point is to prove: Lemma 5.11. For every p 1, the word u p is geodesic in F + sym.

Proof. With the notation of Conjecture 5.3, let T p be the size 2p + 2 tree (10) p 1 . Then T p * u p is defined and equal to T ′ p = 0 p+1 1 p , see Figure 23. We look at the covering relations that are satisfied in T ′ p but not in T p . First, 2p + 1 covers 0 in T ′ p but not in T p . We deduce that every A A A-word transforming T p to T ′ p contains