Christian Bonatti

Sylvain Crovisier

Amie Wilkinson

C 1 -generic conservative diffeomorphisms have trivial centralizer

Keywords: Trivial centralizer, trivial symmetries, C 1 generic properties Centralisateur trivial, symétries triviales, propriétés C 1 -génériques

published or not. The documents may come

Introduction

Let M be a connected compact manifold. The centralizer of a C r diffeomorphism f ∈ Diff r (M ) is defined as Z r (f ) := {g ∈ Diff r (M ) : f g = gf }.

Clearly Z r (f ) always contains the group < f > of all the powers of f . We say that f has trivial centralizer if Z r (f ) =< f >. A diffeomorphism f with trivial centralizer posesses no smooth symmetries, such as those that would arise if, for example, f embedded in a flow or were the lift of another diffeomorphism. S. Smale asked the following:

Question 1 ( [START_REF] Smale | Dynamics retrospective: great problems, attempts that failed. Nonlinear science: the next decade[END_REF][START_REF] Smale | Mathematical problems for the next century[END_REF]). Consider the set of C r diffeomorphisms of a compact connected manifold M with trivial centralizer. 1. Is this set dense in Diff r (M )?

2. Is it residual in Diff r (M )? That is, does it contain a dense G δ subset? 3. Does it contain an open and dense subset of Diff r (M )? 1 We refer to [START_REF] Bonatti | The centralizer of a C 1 generic diffeomorphism is trivial[END_REF] for a discussion on this question. To summarize, we mention some cases in which it has been answered:

• N. Kopell [Ko] solved the one-dimensional smooth case: the set of diffeomorphisms with trivial centralizer contains an open and dense subset of Diff r (S 1 ).

• For r ≥ 2 on higher dimensional manifolds, there exists some results about the set of diffeomorphisms with trivial centralizer under additional dynamical assumptions. For instance, this set contains an open and dense subset among the set of Axiom A diffeomorphisms in Diff ∞ (M ) possessing at least one periodic sink or source [START_REF] Palis | Rigidity of centralizers of diffeomorphisms[END_REF] or defined on a surface [F], and generic among those satisfying the no-cycles condition [F]; and it is locally residual among a class of partially hyperbolic C ∞ diffeomorphisms with 1-dimensional center [START_REF] Burslem | Centralizers of partially hyperbolic diffeomorphisms[END_REF].

• In the C 1 setting, Y. Togawa proved [START_REF] Togawa | Generic Morse-Smale diffeomorphisms have only trivial symmetries[END_REF][START_REF] Togawa | Centralizers of C 1 -diffeomorphisms[END_REF] that generic Axiom A diffeomorphisms have trivial centralizer.

This paper deals with the first two parts of Question 1 for all compact M in the case of volume-preserving and symplectic C 1 diffeomorphisms. If M is a symplectic manifold, then Symp 1 (M ) denotes the space of C 1 symplectomorphisms of M . If M carries a volume µ, then we denote by Diff 1 µ (M ) the space of C 1 diffeomorphisms of M that preserve µ. The spaces Diff 1 (M ), Symp 1 (M ), Diff 1 µ (M ) are Baire spaces in the C 1 topology. Recall that a residual subset of a Baire space is one that contains a countable intersection of open-dense sets.

Theorem A. Let M be a compact, connected manifold of dimension at least 2. Then:

(a) In Symp 1 (M ) the set of diffeomorphisms f whose centralizer Z 1 (f ) ∩ Symp 1 (M ) is trivial is a residual subset. (b) In Diff 1 µ (M ) the set of diffeomorphisms f whose centralizer Z 1 (f ) ∩ Diff 1 µ (M ) is trivial is a residual subset.
In a recent work [START_REF] Bonatti | C 1 -generic diffeomorphisms have trivial centralizer[END_REF] we also solved Questions 1 a) and b) in the C 1 dissipative case: the set of diffeomorphisms f whose centralizer Z 1 (f ) is trivial is a residual subset of Diff 1 (M ). On the other hand with G. Vago [BCVW] we answered negatively to Questions 1 c) in the dissipative and in the symplectic case: for any compact symplectic manifold M , there exists a familly of symplectomorphisms of M with large centraliser Z 1 (f ) ∩ Symp 1 (M ) and that is dense in a non-empty open subset U of Symp 1 (M ); with a different method we also constructed a family of diffeomorphisms of M with large centraliser Z 1 (f ) that is dense in a non-empty open subset U of Diff 1 (M ). The existence of such examples in the volume-preserving setting in dimension larger or equal to 3 remains open.

In [START_REF] Kopell | Commuting diffeomorphisms. In Global Analysis[END_REF][START_REF] Palis | Rigidity of centralizers of diffeomorphisms[END_REF][START_REF] Palis | Centralizers of Anosov diffeomorphisms on tori[END_REF][START_REF] Burslem | Centralizers of partially hyperbolic diffeomorphisms[END_REF][START_REF] Togawa | Generic Morse-Smale diffeomorphisms have only trivial symmetries[END_REF][START_REF] Togawa | Centralizers of C 1 -diffeomorphisms[END_REF], the proof of the triviality of the centralizer splits in two parts: the first implies that, for a generic f and for any diffeomorphism g that commutes with f , we have g = f α on a dense subset of M , where α is a function that is f -invariant; the second part considers the global dynamics and proves that α is bounded, then constant. It is for this second part that additional assumptions such as hyperbolicity are often required. Let us mention that for conservative diffeomorphisms of the 2-sphere, L. Burslem has shown [START_REF] Burslem | Centralizers of area preserving diffeomorphisms on S 2[END_REF] a partial result for the topology C r with r ≥ 16 (more precizely, she obtained the first step).

The proof of theorem A also decomposes in two steps. The first one is provided by the following result.

Theorem B. (a) For any diffeomorphism f in a residual set R ⊂ Diff1 (M ), any periodic point x ∈ P er(f ) is hyperbolic and for any g ∈ Z 1 (f ) there exists n ∈ Z such that g coincides with f n on W s (x). (b) For any diffeomorphism f in a residual set R symp ⊂ Symp 1 (M ), any hyperbolic periodic point x ∈ P er(f ) and any g ∈ Z 1 (f ), there exists n ∈ Z such that g coincides with f n on W s (x). (c) For any diffeomorphism f in a residual set R µ ⊂ Diff 1 µ (M ), any hyperbolic periodic point x ∈ P er(f ) and any g ∈ Z 1 (f ), there exists n ∈ Z such that g coincides with f n on either W s (x) or W u (x).

The second step is provided by the following genericity result about conservative diffeomorphisms. This immediately concludes the proof of theorem A.

Theorem ( [BC, ABC]). For any compact connected manifold M , there are resid-

ual sets R symp ⊂ Symp 1 (M ) and R µ ⊂ Diff 1 µ (M ) such that, every f ∈ R symp ∪ R µ has a hyperbolic periodic point p with W s (p) = W u (p) = M.
More generally, Theorem B naturally applies to the class of C 1 diffeomorphisms satisfying a property we call periodic accessibility: this property says that any two points x, y in a dense subset E ⊂ M of non-periodic points may be joined by a finite sequence x 0 = x, x 1 , . . . , x n = y, x i ∈ E such that for any i ∈ {0, . . . , n -1}, the points x i and x i+1 belong to the closure W s (O i ) of the stable manifold or to the closure W u (O i ) of the unstable manifold of a hyperbolic periodic orbit O i .

Question 2. Is periodic accessibility generic in Diff r (M )?

As a weaker problem 1 , one can also ask if, for generic diffeomorphims, the union of the stable manifolds of the periodic points are dense in M .

As an immediate corollary of Theorem B, it follows that if f ∈ R satisfies the periodic accessibility property, then Z 1 (f ) is trivial. Since the periodic accessibility property is satisfied by Axiom A diffeomorphisms, one can recover in this way Togawa's result that the C 1 generic Axiom A diffeomorphism has trivial centralizer. The proof in the general case [START_REF] Bonatti | The centralizer of a C 1 generic diffeomorphism is trivial[END_REF] is much more delicate and involves new ideas.

Theorem B (a) was previously proved by Togawa [START_REF] Togawa | Generic Morse-Smale diffeomorphisms have only trivial symmetries[END_REF][START_REF] Togawa | Centralizers of C 1 -diffeomorphisms[END_REF]. We present here a variation of his argument, based on an unbounded distortion property that is discussed in section 1; we note below that this proof also handles the "Lipschitz centralizer" and this will be crucial for [START_REF] Bonatti | The centralizer of a C 1 generic diffeomorphism is trivial[END_REF]. When we started with this work we were not aware of Togawa's papers and we used a stronger unbounded distortion property which is much more difficult to obtain. We think that our former approach has independent interest and could have further application: it is written in the first version [START_REF] Bonatti | Centralizers of C 1 -generic diffeomorphisms[END_REF] of this article. The conservative cases (b) and (c) of Theorem B can be derived as for case (a) with an extension result presented in section 2: any perturbation of the dynamics of a symplectomorphism inside the stable manifold of a periodic point can be realized as a perturbation of the dynamics on M .

The unbounded distortion property

Kopell's proof in [Ko] that the set of diffeomorphisms having a trivial centralizer is open and dense in Diff r (S 1 ) for r ≥ 2 uses the fact that a C 2 diffeomorphism f of [0, 1] without fixed points in (0, 1) has bounded distortion, meaning: for any x, y ∈ (0, 1), the ratio

|f n′ (x)| |f n ′ (y)| (1)
is bounded, independent of n and uniformly for x, y lying in a compact set. A bounded distortion estimate lies behind many results about C 2 , hyperbolic diffeomorphisms of the circle and codimension-1 foliations. Suppose that r ≥ 2. Since Morse-Smale diffeomorphisms are open and dense in Diff r (S 1 ), the proof that the set of diffeomorphisms having a trivial centralizer is open and dense in Diff r (S 1 ) essentially reduces to showing that (C r open and densely) a C r diffeomorphism f : [0, 1] → [0, 1] without fixed points in (0, 1) has trivial centralizer. The bounded distortion of such an f forces its centralizer to embed simultaneously in two smooth flows containing f , one determined by the germ of f at 0, and the other by the germ at 1; for an open and dense set of f ∈ Diff r + [0, 1], these flows agree only at the iterates of f . The r ≥ 2 hypothesis is clearly necessary for bounded distortion.

The central observation and starting point of this paper is that the centralizer of a C 1 diffeomorphism of [0, 1] with unbounded distortion is always trivial. We elaborate a bit on this. Notice that if x and y lie on the same f -orbit, then the ratio in (1) is bounded, independent of n. We show that, C 1 generically among the diffeomorphisms of [0, 1] without fixed points (0, 1), the ratio (1) is uniformly bounded in n only if x and y lie on the same orbit; that is, for a residual set of f , and for all x, y ∈ (0, 1), if

x / ∈ O f (y) = {f n (y) | n ∈ Z}, then lim sup n→∞ |f n′ (x)| |f n ′ (y)| = ∞. (2)
Assume that this unbounded distortion property holds for f . Fix x ∈ (0, 1). A simple application of the Chain Rule shows that if gf = f g, then the distortion in (1) between x and y = g(x) is bounded; hence x and g(x) must lie on the same f -orbit. From here, it is straightforward to show that g = f n , for some n (see Lemma 1.2 below). As in [Ko], a small amount of additional work shows that a residual set in Diff 1 (S 1 ) has trivial centralizer. The bulk of this section is devoted to formulating and proving a higherdimensional version of the argument we have just described. The interval is replaced by an invariant manifold (stable or unstable) of a periodic point. The derivative f ′ in (2) is replaced by the Jacobian of f along the invariant manifold.

Unbounded distortion along stable manifolds of hyperbolic periodic points

Denote by Lip(M ) is the set of lipeomorphisms of M : these are the homeomorphisms g of M such that g and g -1 are Lipschitz. For any hyperbolic periodic orbit O ⊂ M of a diffeomorphism f ∈ Diff 1 (M ) and for x ∈ W s (O) we denote by Jac s (f )(x) the Jacobian of the map induced by

T x f between T x W s (O) and T f (x) W s (O). Definition 1.1. A hyperbolic periodic orbit O ⊂ M of a diffeomorphism f ∈ Diff 1 (M )
(f n )(x)) -log(Jac s (f n )(y))| = ∞.
As mentioned in the previous subsection, unbounded distortion forces trivial centralizers along the stable manifold: Lemma 1.2. Let f ∈ Diff 1 (M ) be a diffeomorphism and O be a hyperbolic periodic orbit having the stable manifold unbounded distortion property. Then for any g ∈

Lip(M ) such that g • f = f • g and g(O) = O, and for any connected component W of W s (O) \ O, there exists an i ∈ Z such that g = f i on W . Remark: Since O is hyperbolic, if g is differentiable at p, then there exists i ∈ Z such that g = f i on W s (O).
Proof. Let D ⊂ W s (O) be the dense set as in definition 1.1.

We claim that: For every x ∈ D \ O, the map g preserves the f -orbit of x. In order to prove it, we note that for any n ∈ N, the relation

gf n = f n g implies Jac s (g)(f n x) Jac s (g)(x) = Jac s (f n )(gx) Jac s (f n )(x) .
Since f n (x) lies in a compact region of W s (O) for all n ∈ N, the left hand side of this expression is uniformly bounded in n. Hence if y = g(x), then the quantity |log(Jac

s (f n )(x)) -log(Jac s (f n )(y))| is bounded in n. Since g(O) = O the point y = g(x) belongs to W s (O)
\ O and the stable manifold unbounded distortion property implies that y belongs to the f -orbit of x, proving the claim.

For i ∈ Z we consider the closed subset

P i = {x ∈ W s (O) \ O, g(x) = f i (x)} of W s (O) \ O.
The claim above shows that their union P contains D, hence is dense in W s (O) \ O. Moreover for i = j the intersection P i ∩ P j is empty (any point in the intersection would be j -i-periodic).

We claim that: For every i ∈ Z, the set P i is open in P . Consider a point x ∈ P i and a neighborhood U ⊂ W s (O) of x that is disjoint from all its iterates. Consider also a point y ∈ P ∩ U close to x. Since g(x) belongs to f i (U ), this is also the case for g(y). On the other hand there exists j ∈ Z such that g(y) = f j (y). Since the iterates of U are pairwise disjoint, it follows that j = i. The map g hence agrees with f i on a neighborhood of x in P ∩ W s (O), proving the claim.

Let us consider a non-empty set P i . Since P is dense in W s (O), it follows from the second claim that the interior of P i in W s (O) is a non-empty f -invariant set in which g = f i . Since O is a hyperbolic periodic orbit, the Lipschitz constant of f i in a neighborhood of O is arbitrarily large when |i| goes to infinity. Using that g is uniformly Lipschitz, one deduces that only a finite number of sets P i are non-empty. Consequently W s (O) \ O is the disjoint union of finitely many closed sets P i . Hence

any connected component W of W s (O) \ O is contained in a single set P i , proving that g = f i on W .
The periodic orbits of a generic diffeomorphism f are fixed by C(f ).

Lemma 1.3. For any diffeomorphism f in a residual set R per of Diff 1 (M ) (resp. of Symp 1 (M ), of Diff 1 µ (M )), for any periodic orbit O of f and for any g

∈ Lip(M ) such that f • g = g • f , we have g(O) = O.
Proof. If f and g commute, then any periodic orbit O of f is sent by g on a periodic orbit O ′ of f having the same period. Since g is Lipschitz, the eigenvalues of O and O ′ for f must coincide. The set R per of diffeomorphisms whose periodic orbits have different eigenvalues hence satisfies the conclusion of the lemma.

To prove Theorem B, it thus remains to show: Proposition 1.4. For any diffeomorphism f in a residual set R ⊂ Diff 1 (M ), any periodic orbit of f is hyperbolic and has the stable manifold unbounded distortion property.

For any diffeomorphism f in a residual set R symp ⊂ Symp 1 (M ), any hyperbolic periodic orbit of f has the stable unbounded distortion property.

For any diffeomorphism f in a residual set R µ ⊂ Diff 1 µ (M ) any hyperbolic periodic O of f whose stable manifold W s (O) has codimension at least dim(M )/2 has the stable manifold unbounded distortion property.

Reduction to contractions of R d

Let B d denote the unit closed ball B(0, 1) of R d and consider the Banach space of C 1 maps B d → R d that send 0 to 0, endowed with the C 1 topology given by the C 1 norm: Let f be a diffeomorphism of a manifold M , p be a periodic point of f , d s be its stable dimension and τ be its period. A stable chart for p is a local chart ψ : R d → M such that, denoting by π s the projection of R d onto the d s first coordinates, we have the following properties.

f -g 1 = sup x∈B d f (x) -g(x) + D x f -D x g
• The domain ψ(R d ) contains p.

• In the chart ψ, the local stable manifold of p contains the graph of a C 1 map g : R d s → R d-d s .

• Let v be equal to π s (ψ -1 (p)) and let θ be the C 1 map defined on a neighborhood of 0 by projecting on the space R d s the dynamics of f τ in the local stable manifold of p:

θ : x → π s • ψ -1 • f τ • ψ(x + v, g(x + v)) -v.
Then, θ belongs to C d s .

The following property is easy to check.

Lemma 1.5. Any hyperbolic periodic point p of a diffeomorphism f has a stable chart ψ. Moreover, for any diffeomorphism g in a C 1 neighborhood U of f , the continuation p g of p also admits the chart ψ as a stable chart. The family of contractions θ g associated to the periodic point p g and to the chart ψ induces a continuous map Θ : U → C d s . This map is open.

In the conservative setting, the same property holds, but the proof is more delicate and is postponed until section 2.

Theorem 1.6. Let Θ : U → C d s be a family of contractions associated to a periodic point p and a stable chart ψ as in lemma 1.5. Then the map Θ :

U ∩Symp 1 (M ) → C d s is open.
If the dimension d s of the stable space of p is greater than or equal to dim(M )/2, then the map Θ :

U ∩ Diff 1 µ (M ) → C d s is open.
The major ingredient in the proof of Proposition 1.4 is the following.

Proposition 1.7. There is a residual set S d ⊂ C d and a dense set D ⊂ B d \ {0} such that any f ∈ S d has the following property: for any points x ∈ D, y ∈ B d \ {0} not in the same orbit and for any N ≥ 1 there exists n ≥ 1 such that

|log(Jac(f n )(x)) -log(Jac(f n )(y))| > N. (3)
All these results together allow us to prove that C 1 generically, the stable manifold unbounded distortion property holds.

Proof of Proposition 1.4. For any integer n ≥ 0, there exists • a family P n of pairwise disjoint open subsets whose union is dense in Diff 1 (M ),

• for each U ∈ P n , finitely many charts ψ 1 , . . . , ψ s : R d → M , such that any diffeomorphism f ∈ U has the following properties:

• f has s periodic points of period less than n, all are hyperbolic. Each domain ψ i (R d ) contains exactly one of them, it is called p i,f and its stable dimension is denoted by d s i and its period by τ i .

• The chart ψ i is a stable chart for p i,f . By Lemma 1.5, the chart ψ i induces an open map Θ i : U → C d s i , so that Θ -1 i (S d s i ) is residual in U where S d s i is the residual subset of C d s i provided by proposition 1.7. Suppose that f belongs to this residual set. The dynamics of f τ i on the local stable manifold of p i are differentiably conjugate to the dynamics of a map Θ i (f ) ∈ S d s i , proving that p i has the stable unbounded distortion property.

The union R n over U ∈ P n of the sets s i=1 Θ -1 i (S d s i ) is residual in Diff 1 (M ) and Proposition 1.4 holds with the residual set R = ∩ n R n . The proof in the conservative cases is similar and uses Theorem 1.6.

Huge distortion: proof of Proposition 1.7

We fix a countable dense subset D of Int(B d ) \ {0} and a countable basis of compact neighborhoods B for B d \ {0}. For any x ∈ D and any compact set Λ ∈ B, the set of contractions f ∈ C d such that the orbits of x and of Λ are disjoint is an open set O(x, Λ). For any N ≥ 1, the subset O(x, Λ, N ) ⊂ O(x, Λ) of contractions f such that for any y ∈ Λ we have ( 3) is open. Thus the set

S d = x,Λ,N O(x, Λ, N ) ∪ (C d \ O(x, Λ))
is a G δ and any diffeomorphism f ∈ S d satisfies the conclusion of Proposition 1.7. In order to prove Proposition 1.7 we must show that S d is dense (hence residual) in C d . This is a consequence of the following perturbation result.

Lemma 1.8. The set O(x, Λ, N ) is dense in O(x, Λ)
Proof. Consider a contraction f ∈ O(x, Λ). By performing C 1 small perturbation, we can assume that f is linear in a neighborhood V of 0. Now consider i 0 ≥ 0 such that all the iterates f i (x) and f i (Λ) with i ≥ i 0 are contained in V . There exists a C 1 small perturbation with compact support g 1 of the linear map D 0 f such that Jacg 1 (0) = Jacf (0).

For some n ≥ i 0 and any i ∈ {i 0 , . . . n}, the map f agrees with D 0 f in a neighborhood of f i (x) and can be replaced by the map

g i,ε : z → f i (x) + ε.g 1 z -f i (x) ε .
By choosing ε small enough, all these perturbations have disjoint support. We thus obtain a map g that is still C 1 close to f . Note that f and g agree on the positive orbit of x and of a neighborhood of K.

We have for any point y ∈ Λ,

|log(Jac(f n )(x)) -log(Jac(f n )(y))| ≥ (n -i 0 ) |log Jacg 1 (0) -log Jacf (0)| -log(Jac(f i 0 )(x)) -log(Jac(f i 0 )(y)) ,
which is larger than N , if n has been chosen large enough. Hence, g belongs to O(x, Λ, N ).

Conservative extension results

We explain in this part how a perturbation of a conservative diffeomorphism along a submanifold W can be extended to a conservative perturbation on the whole manifold M . This implies Theorem 1.6: the results proven in this section will be applied to the case where W is an invariant manifold of a hyperbolic periodic point p. In the volume-preserving case, we assume that dim(W ) ≤ 1 2 dim(M ) (note that this hypothesis is always satisfied either by the stable or by the unstable manifold of p). In the symplectic case, there is no additional hypothesis, but we use the following well-known fact.

Lemma 2.1. Let f ∈ Symp 1 (M ) and let p be a hyperbolic periodic point for f . Then W s (p) and W u (p) are Lagrangian submanifolds of M .

Proof. Let x ∈ W s (p), and let v, w ∈ T x W s (p) be tangent vectors to W s (p). On the one hand, since f is a symplectomorphism, we have

ω(D x f k (v), D x f k (w)) = ω(v, w),
for all k ∈ Z. On the other hand, as k → +∞, we have

ω(D x f k (v), D x f k (w)) → 0.
Hence ω vanishes identically on W s (p). The same is true for W u (p). Since W s (p) and W u (p) have complementary dimension and ω is nondegenerate, they must have the same dimension. Hence, both are Lagrangian submanifolds of M .

The symplectic case

Proposition 2.2. Let M be a symplectic manifold and z a point contained in a C 1 Lagrangian submanifold W ⊂ M . Then there exists in W a disk D = B W (z, r 0 ) centered at z such that, for every neighborhood U ⊂ M of D and every ε > 0, there exists δ > 0 with the following property. ). The natural thing to try to do is to set ϕ = ψ * in a neighborhood of the 0-section, symplectically interpolating between ψ * and Id using a generating function. This simple approach fails, however, because ψ is only C 1 , and so ψ * is merely continuous.

(Even assuming that ψ is C 2 does not help: in order to control the C 1 size of such a map, it is necessary to have some control on the C 2 size of ψ, and we cannot assume any such control). Using a convolution, it is possible to overcome this problem. This approach mirrors that in [BGV], but in the symplectic setting. The problem is local and one can work in R 2n endowed with the standard symplectic form ω = i du i ∧ dv i where u = (u 1 , . . . , u n ), v = (v 1 , . . . , v n ). By a symplectic change of coordinates, we may assume that the disk D lies inside a disk {(u, v), u ≤ R, v = 0}. We define ψ using a generating function S.

We first recall the definition and properties of generating functions. Suppose that h : R 2n → R 2n is a C r symplectomorphism, taking the form:

h(u, v) = (ξ(u, v), η(u, v)),
with ξ, η : R 2n → R n and h(0, 0) = (0, 0). Assume that the partial derivative matrix ∂ ∂v η(u, v) is invertible (this is the case for instance if h preserves R n × {0}). We can solve for η = η(u, v) to obtain new coordinates (u, η) on a small neighborhood of (0, 0) in R 2n . Since h is symplectic, the 1-form α = i v i du i + ξ i dη i is closed, and hence, exact. Thus there exists a C r+1 function S = S(u, η), unique up to adding a constant, defined in a neighborhood of (0, 0), such that dS = α. The function S is called a generating function for h.

On the other hand, any C r+1 function S = S(u, η) satisfying the nondegeneracy condition that ∂ 2 ∂u∂η S is everywhere nonsingular is the generating function of a C r symplectic diffeomorphism. Solving for α in the equation

dS = ∂S ∂u du + ∂S ∂η dη = α = vdu + ξdη,
we obtain the system:

∂S ∂u = v; ∂S ∂η = ξ.
The nondegeneracy condition implies that this system can be solved implicitly for a C r function η = η(u, v). We then obtain a C r symplectomorphism:

h(u, v) = ∂S ∂η (u, η(u, v)), η(u, v) ,
and S is a generating function for h.

It is easy to see that the generating function for the identity map is

S 0 (u, η) = u • η = n i=1 u i η i .
Claim 1. For every ε > 0, there exists δ > 0 such that, if d C 2 (S, S 0 ) < δ then d C 1 (h, Id) < ε.

Proof. This follows from the implicit function theorem, and the details are omitted.

Returning to the proof of Proposition 2.2, assume that ψ :

D → D is written in u-coordinates as ψ(u 1 , . . . , u n ) = (ψ 1 (u 1 , . . . , u n ), . . . , ψ n (u 1 , . . . , u n )).
We may assume that the domain of ψ has been extended to R n . To prove Proposition 2.2, it suffices to prove the following lemma.

Lemma 2.3. Given a disk D ⊂ R n , and a neighborhood U of D × {0} in R 2n , there exists C > 0 with the following property.

For every C 1 diffeomorphism ψ : R n → R n , equal to the identity on a neighborhood of ∂D, there is a C 2 function S : R 2n → R such that:

1. d C 2 (S 0 , S) ≤ Cd C 1 (ψ, Id),
2. S = S 0 outside of U , 3. ∂S ∂u (u, 0) = 0 for all u ∈ R and 4. ∂S ∂η (u, 0) = ψ(u) for all u ∈ D. Note that condition 1. implies that S is nondegenerate, provided that d C 1 (ψ, Id) is sufficiently small. Proof of Lemma 2.3. To illustrate the argument in a simple case, we first prove the lemma for n = 1. The proof of the general case is very similar. Let

a(u) = ψ ′ (u) -1. Note that a is a continuous map, a ∞ ≤ d C 1 (ψ, Id), and a(u) = 0 if u / ∈ int(D). Let Φ : R → [0, 1] be a C ∞ function satisfying:
• Φ(0) = 1 and Φ = 0 outside of (-1, 1),

• Φ (k) (0) = 0, for all k ≥ 1, • R Φ(w) dw = 1. Fix a point u * ∈ ∂D, so that ψ(u * ) = u * . For (u, η) ∈ R 2 , η = 0, let: Q(u, η) = η u u * R Φ(w) a(x -wη) dw dx.
For η = 0, one can make the change of variables w ′ = x -wη and get

Q(u, η) = Sign(η) u u * R Φ
x -w ′ η a(w ′ ) dw ′ dx.

Let ρ : R 2 → R be a C ∞ bump function identically equal to 1 on a neighborhood of D × {0} and vanishing outside of U . Consider S = S 0 + ρ Q.

Lemma 2.3 in the case n = 1 is a direct consequence of:

Claim 1. The map Q : R 2 → R is C 2 and there is C = C(U ) > 0 such that: Proof. We repeat the calculations from the proof of Lemma 1 in the general setting. When t = 0, the change of variable w ′ = u -tw gives α ⋆t (u) = Sign(t)

1. Q | U C 2 ≤ C a ∞ ,
R n Φ n u -tw ′ α(w ′ ) dw ′ , d dt α ⋆t (u) = - R n
(dΦ n (w).w + (n -1)Φ n (w)) α(u -tw) dw.

One deduces:

∂Q ∂u = n i=1 η i R n Φ n (w) α i (u -tw) dw = n i=1 η i |η i | n R n Φ n u -w ′ η i α i (w ′ ) dw ′ , ∂Q ∂η i = u u * d dη i α ⋆η i i = - R n
(dΦ n (w).w + (n -1)Φ n (w))

x=u-η i w x=u * -η i w α i (w) dw,

∂ 2 Q ∂u ∂η i = - R n
(dΦ n (w).w + (n -1)Φ n (w)) α i (u -η i w) dw,

∂ 2 Q ∂u 2 = n i=1 R n
dΦ n (w) α i (u -η i w) dw, and finally:

∂ 2 Q ∂η i ∂η j = δ i,j R n
(dΦ n (w).w + (n -1)Φ n (w)) [α i (x -η i w).w] x=u x=u * dw.

It is not difficult to verify that 1.-3. hold.

The proof of Lemma 2.3 is now complete.

The volume-preserving case

Proposition 2.4. Let M be a Riemannian manifold endowed with a volume form µ and W be a C 1 submanifold satisfying dim(W ) ≤ codim(W ).

Centered at any point z ∈ W , there exists a disk D = B W (z, r 0 ) of W such that, for every neighborhood U ⊂ M containing D and every ε > 0, there exists δ > 0 with the following property.

For every C 1 diffeomorphism ψ : D → D satisfying:

a. ψ = Id on a neighborhood of ∂D, and

  . The set of embeddings B d → R d fixing 0 defines an open subset that will be denoted by D d . A contraction of R d is an element of D d that sends B d into B(0, 1), so that 0 is a (hyperbolic) sink that attracts all the points in B d . The set of contractions of R d is an open subset C d ⊂ D d , and hence a Baire space.

For every C 1

 1 diffeomorphism ψ : D → D satisfying: a. ψ = Id on a neighborhood of ∂D, and b. d C 1 (ψ, Id) < δ, there exists ϕ ∈ Symp 1 (M ) such that: 1. ϕ = Id on M \ U , 2. ϕ = ψ on D, and 3. d C 1 (ϕ, Id) < ε. Proof. The basic strategy is first to symplectically embed the disk D as the 0-section of its cotangent bundle T * D. On T * D, the symplectic form is ω = dα, where α is the canonical one-form on T * D. Any diffeomorphism ψ : D → D lifts to a canonical symplectomorphism ψ * : T * D → T * D; namely the pull-back map (ψ, Dψ -1

ΦΦ

  2. ∂Q∂u (u, 0) = 0, for all u ∈ R, and3. ∂Q ∂η (u, 0) = u u * a(x) dx = ψ(u) -u, for all u ∈ R. Proof. We derive explicitly the formulas: ′ x -w ′ η (x -w ′ ) a(w ′ ) dw ′ dx = -u u * R Φ ′ (w) w a(x -wη) dw dx = -R Φ ′ (w) w u-wη u * -wη a(x ′ ) dx ′ dw, ∂ 2 Q ∂η∂u = -R Φ ′ (w) w a(u -wη) dw, ∂ 2 Q ∂u 2 = 1 |η| R Φ ′ u -w ′ η a(w ′ ) dw ′ = R Φ ′ (w) a(u -wη) dw, ′(w) w 2 (a(u -wη) -a(u * -wη)) dw.

  has the stable manifold unbounded distortion property if there exists a dense set D ⊂ W s (O) with the following property: for any points x ∈ D and y ∈ W

s (O) \ O not in the same f -orbit and for any N ≥ 1 there exists n ≥ 1 such that |log(Jac s

This problem is often proposed as a conjecture by the first author.

Properties 1. and 2. follow immediately from these formulas. To see 3., note that We now turn to the case n ≥ 1 in Lemma 2.3. For i = 1, . . . n, let α i be the continuous 1-form defined by

where π i : R n → R is the projection onto the ith coordinate. As above, fix a point u * ∈ ∂D, so that ψ(u * ) = u * . Then we have the formula:

where the right-hand side is a path integral evaluated on any path from u * to u = (u 1 , . . . , u n ). Furthermore, we have

, where a(u) = ψ ′ (u) -1, as above. Let Φ n : R n → R be an n-dimensional bell function:

For each 1-form α, and t ∈ R, we define a new 1-form α ⋆t i on R n by taking the convolution:

We integrate along any path from u * to u and set

This is well-defined since u u * α i (u -tw) is independent of choice of path. Let ρ n : R 2n → [0, 1] be a C ∞ bump function vanishing identically outside of U and equal to 1 on a neighborhood of D. As before, the map S = S 0 + ρ n Q satisfies the conclusions of Lemma 2.3 provided the following claim holds.

Proof. Let n = dim(M ). By a local change of coordinates, we may assume that µ is the standard volume form dx 1 ∧ • • • ∧ dx n on a neighborhood of the origin in R n . By composing these coordinates with an isometry of R n , we may further assume that D is the graph of a

preserves volume. Applying this change of coordinates, we may assume that D lies in the coordinate plane

that agrees with ψ on D. This symplectomorphism is C 1 isotopic to the identity through symplectomorphisms {ϕ t } t∈[0,1] , where ϕ 1 = Id (to obtain this isotopy, just choose a smooth isotopy of the generating function for ψ to the generating function for the identity). Now we extend ϕ 0 to R n using this isotopy to obtain a locally-supported volumepreserving diffeomorphism that agrees with ψ on D. More precisely, choose an appropriate C ∞ bump function ρ : R n-2k → [0, 1], and set ϕ(x 1 , . . . , x n ) = (ϕ ρ( (x 2k+1 ,...,xn) ) (x 1 , . . . , x 2k ), x 2k+1 , . . . , x n ). This is the desired map ϕ.


Diego Arcis

Luis Paris

Ordering Garside groups

We introduce a structure on a Garside group that we call Dehornoy structure and we show that an iteration of such a structure leads to a left-order on the group. We define two conditions on a Garside group G and we show that, if G satisfies these two conditions, then G has a Dehornoy structure. Then we show that the Artin groups of type A and of type I 2 (m), m ≥ 4, satisfy these conditions, and therefore have Dehornoy structures. As indicated by the terminology, one of the orders obtained by this method on the Artin groups of type A coincides with the Dehornoy order. 20F36

Introduction

A group G is said to be left-orderable if there exists a total order < on G invariant by left-multiplication. Recall that a subset P of G is a subsemigroup if αβ ∈ P for all α, β ∈ P. It is easily checked that a left-order < on G is determined by a subsemigroup P such that G = P ⊔ P -1 ⊔ {1}: we have α < β if and only if α -1 β ∈ P. In this case the subsemigroup P is called the positive cone of <.

The first explicit left-order on the braid group B n was determined by Dehornoy [START_REF] Dehornoy | Braid groups and left distributive operations[END_REF]. The fact that B n is left-orderable is important, but, furthermore, the Dehornoy order is interesting by itself, and there is a extensive literature on it. We refer to Dehornoy-Dynnikov-Rolfsen-Wiest [START_REF] Dehornoy | Ordering braids[END_REF] for a complete report on left-orders on braid groups and on the Dehornoy order in particular. The definition of the Dehornoy order is based on the following construction.

Let G be a group and let S = {s 1 , s 2 , . . . , s n } be a finite ordered generating set for G. Let i ∈ {1, 2, . . . , n}. We say that α ∈ G is s i -positive (resp. s i -negative) if α is written in the form α = α 0 s i α

1 • • • s i α m (resp. α = α 0 s -1 i α 1 • • • s -1 i α m ) with m ≥ 1
and α 0 , α 1 , . . . , α m ∈ s i+1 , . . . , s n . For each i ∈ {1, 2, . . . , n} we denote by P + i (resp. P - i ) the set of s i -positive elements (resp. s i -negative elements) of G. The key point in the definition of the Dehornoy order is the following.

Theorem 1.1 (Dehornoy [START_REF] Dehornoy | Braid groups and left distributive operations[END_REF]) Let G = B n+1 be the braid group on n + 1 strands and let S = {s 1 , s 2 , . . . , s n } be its standard generating set. For each i ∈ {1, 2, . . . , n} we have the disjoint union s i , s i+1 , . . . , s n = P + i ⊔ P - i ⊔ s i+1 , . . . , s n .

Let G = B n+1 be the braid group on n + 1 strands. Set P D = P + 1 ⊔ P + 2 ⊔ • • • ⊔ P + n . Then, by Theorem 1.1, P D is the positive cone for a left-order < D on G. This is the Dehornoy order.

A careful reader will notice that Theorem 1.1 leads to more than one left-order on B n+1 . Indeed, if ǫ = (ǫ 1 , ǫ 2 , . . . , ǫ n ) ∈ {+, -} n , then P ǫ = P ǫ 1 1 ⊔ P ǫ 2 2 ⊔ • • • ⊔ P ǫn n is a positive cone for a left-order on B n+1 . The case ǫ = (+, -, +, . . . ) is particularly interesting because, by Dubrovina-Dubrovin [START_REF] T V Dubrovina | On braid groups[END_REF], in this case P ǫ determines an isolated left-order in the space of left-orders on B n+1 .

Our goal in the present paper is to extend the Dehornoy order to some Garside groups.

A first approach would consist on keeping the same definition, as follows. Let G be a group and let S = {s 1 , s 2 , . . . , s n } be a finite ordered generating set for G. Again, we denote by P + i (resp. P - i ) the set of s i -positive elements (resp. s i -negative elements) of G. Then we say that S determines a Dehornoy structure (in Ito's sense) if, for each i ∈ {1, . . . , n}, we have the disjoint union s i , s i+1 , . . . , s n = P + i ⊔P - i ⊔ s i+1 , . . . , s n . In this case, as for the braid group, for each ǫ ∈ {+, -} n the set P ǫ = P ǫ 1 1 ⊔P ǫ 2 2 ⊔• • •⊔P ǫn n is the positive cone for a left-order on G. This approach was used by Ito [START_REF] Ito | Dehornoy-like left orderings and isolated left orderings[END_REF] to construct isolated left-orders in the space of left-orders of some groups.

In the present paper we will consider another approach of the Dehornoy order in terms of Garside groups (see Dehornoy [START_REF] Dehornoy | Alternating normal forms for braids and locally Garside monoids[END_REF], Fromentin [START_REF] Fromentin | Every braid admits a short sigma-definite expression[END_REF], Fromentin-Paris [START_REF] Fromentin | A simple algorithm for finding short sigma-definite representatives[END_REF]), and our definition of Dehornoy structure will be different from that in Ito's sense given above.

In Section 2 we recall some basic and preliminary definitions and results on Garside groups. We refer to Dehornoy et al. [START_REF] Dehornoy | Foundations of Garside theory[END_REF] for a full account on the theory. In Section 3 we give our (new) definition of Dehornoy structure and show how such a structure leads to a left-order on the group (see Proposition 3.1). Then we define two conditions on a Garside group, that we call Condition A and Condition B, and show that a Garside group which satisfies these two conditions has a Dehornoy structure (see Theorem 3.2).

The aim of the rest of the paper is to apply Theorem 3.2 to the Artin groups of type A, that is, the braid groups, and the Artin groups of dihedral type. In Section 4 we a = u p • • • u 2 u 1 of a over S satisfying (u p • • • u i ) ∧ R ∆ = u i for all i ∈ {1, . . . , p}. We define the left greedy normal form of an element of M in a similar way. The following two theorems contain several key results of the theory of Garside groups.

Theorem 2.1 (Dehornoy-Paris [START_REF] Dehornoy | Gaussian groups and Garside groups, two generalisations of Artin groups[END_REF], Dehornoy [5]) [START_REF] Brieskorn | Artin[END_REF] Let a ∈ M and let a = u p • • • u 2 u 1 be the greedy normal form of a. Then lg(a) = p.

(

) Let α ∈ G. There exists a unique pair (a, b) ∈ M × M such that α = ab -1 and a ∧ R b = 1. 2
In that case we have lg(α) = lg(a) + lg(b).

The expression of α given in Theorem 2.1 ( 2) is called the (right) orthogonal form of α. The left orthogonal form of an element of G is defined in a similar way.

We say that an element

a ∈ M is unmovable if ∆ ≤ R a or, equivalently, if ∆ ≤ L a.
Theorem 2.2 (Dehornoy-Paris [START_REF] Dehornoy | Gaussian groups and Garside groups, two generalisations of Artin groups[END_REF], Dehornoy [5]) Let α ∈ G. There exists a unique pair (a, k) ∈ M × Z such that a is unmovable and α = a∆ k .

The expression of α given above is called the (right) ∆-form of α. We define the left ∆-form of an element of G in a similar way.

Definition Let δ be a balanced element of M . Denote by G δ (resp. M δ ) the subgroup of G (resp. the submonoid of M ) generated by Div(δ). We say that (G δ , M δ , δ) is a parabolic substructure of (G, M, ∆) if δ is balanced and Div(δ

) = Div(∆) ∩ M δ . In that case G δ is called a parabolic subgroup of G and M δ is called a parabolic submonoid of M .
Remark Let H be a parabolic subgroup of G. Then there exists a unique parabolic substructure (G δ , M δ , δ) of (G, M, ∆) such that H = G δ . Indeed, the above element δ should be the greatest element in H ∩ Div(∆) for the order relation ≤ R , hence δ is entirely determined by H . Similarly, if N is a parabolic submonoid of M , then there exists a unique parabolic substructure (G δ , M δ , δ) such that N = M δ , where δ is the greatest element of Div(∆) ∩ N for the order relation ≤ R . So, we can speak of a parabolic subgroup or of a parabolic submonoid without necessarily specifying the corresponding element δ or the triple (G δ , M δ , δ).

Theorem 2.3 (Godelle [START_REF] Godelle | Parabolic subgroups of Garside groups[END_REF]) Let (H, N, δ) be a parabolic substructure of (G, M, ∆).

(1) H is a Garside group with Garside structure (H, N, δ).

(2) Let a ∈ N and let a = u p • • • u 2 u 1 be the greedy normal form of a with respect to (G, M, ∆). Then u i ∈ Div(δ) for all i ∈ {1, 2, . . . , p} and a = u p • • • u 2 u 1 is the greedy normal form of a with respect to (H, N, δ).

(

) Let α, β ∈ H and γ ∈ G such that α ≤ R γ ≤ R β . Then γ ∈ H . (4) Let α, β ∈ H . Then α ∧ R β, α ∨ R β ∈ H . 3
(5) Let α ∈ H and let α = ab -1 be the orthogonal form of α with respect to (G, M, ∆). Then a, b ∈ N and α = ab -1 is the orthogonal form of α with respect to (H, N, δ).

Example Let S be a finite set. A Coxeter matrix over S is a square matrix M = (m s,t ) s,t∈S indexed by the elements of S with coefficients in N ∪ {∞} such that m s,s = 1 for all s ∈ S and m s,t = m t,s ≥ 2 for all s, t ∈ S, s = t. If s, t are two letters and m is an integer ≥ 2 we denote by Π(s, t, m) the word sts

• • • of length m. In other words Π(s, t, m) = (st) m 2 if m is even and Π(s, t, m) = (st) m-1
2 s if m is odd. The Artin group associated with M is the group A = A M defined by the presentation A = S | Π(s, t, m s,t ) = Π(t, s, m s,t ) for s, t ∈ S, s = t and m s,t = ∞ .

The Coxeter group associated with M is the quotient W = W M of A by the relations s 2 = 1, s ∈ S. We say that A is of spherical type if W is finite. The braid groups are the star examples of Artin groups of spherical type.

We denote by A + the monoid having the following monoid presentation.

A + = S | Π(s, t, m s,t ) = Π(t, s, m s,t ) for s, t ∈ S, s = t and m s,t = ∞ + .
By Paris [START_REF] Paris | Artin monoids inject in their groups[END_REF] the natural homomorphism A + → A is injective. So, we can consider A + as a submonoid of A. It is easily checked that A + ∩ (A + ) -1 = {1}, hence we can consider the order relations ≤ R and ≤ L on A. Suppose that A is of spherical type. Then, by Brieskorn-Saito [START_REF] Brieskorn | Artin[END_REF] and Deligne [START_REF] Deligne | Les immeubles des groupes de tresses généralisés[END_REF], for all α, β ∈ A the elements α ∧ R β and α ∨ R β exist, and (A, A + , ∆) is a Garside structure, where ∆ = ∨ R S. Let X be a subset of S and let A X be the subgroup of A generated by X . Then, again by Brieskorn-Saito [START_REF] Brieskorn | Artin[END_REF] and Deligne [START_REF] Deligne | Les immeubles des groupes de tresses généralisés[END_REF], A X is a parabolic subgroup of A and it is an Artin group of spherical type.

The triple (G, M, ∆) denotes again an arbitrary Garside structure on a group G. Besides the greedy normal forms, we will use some other normal forms of the elements of M defined from a pair (N 2 , N 1 ) of parabolic submonoids of M . Their definition is based on the following. Proposition 2.4 (Dehornoy [START_REF] Dehornoy | Alternating normal forms for braids and locally Garside monoids[END_REF]) Let N be a parabolic submonoid of M . For each a ∈ M there exists a unique

b ∈ N such that {c ∈ N | c ≤ R a} = {c ∈ N | c ≤ R b}.
The element b of Proposition 2.4 is called the (right) N -tail of a and is denoted by b = τ N (a) = τ N,R (a). We define in a similar way the left N -tail of a, denoted by τ N,L (a). Now, assume that N 1 and N 2 are two parabolic submonoids of M such that N 2 ∪ N 1 generates M . Then each nontrivial element a ∈ M is uniquely written in the form

a = a p • • • a 2 a 1 where a p = 1, a i = τ N 1 (a p • • • a i ) if i is odd, and a i = τ N 2 (a p • • • a i ) if i is
even. This expression is called the (right) alternating form of a with respect to (N 2 , N 1 ). Note that we may have a 1 = 1, but a i = 1 for all i ∈ {2, . . . , p}. The number p is called the (N 2 , N 1 )-breadth of a and is denoted by p = bh(a) = bh N 2 ,N 1 (a). By extension we set bh(1) = 1 so that a ∈ N 1 ⇔ bh(a) = 1. Now, consider the standard Garside structure (B n+1 , B + n+1 , ∆) on the braid group B n+1 . Let S = {s 1 , s 2 , . . . , s n } be the standard generating system of B n+1 , N 1 be the submonoid of B + n+1 generated by {s 2 , . . . , s n }, and N 2 be the submonoid generated by {s 1 , . . . , s n-1 }. Then N 1 and N 2 are parabolic submonoids of B + n+1 and they are both isomorphic to B + n . Observe that N 1 ∪ N 2 generates B + n+1 , hence we can consider alternating forms with respect to (N 2 , N 1 ). The definitions of the next section are inspired by the following.

Theorem 2.5 (Fromentin-Paris [START_REF] Fromentin | A simple algorithm for finding short sigma-definite representatives[END_REF]) Let a ∈ B + n+1 and k ∈ Z. Then ∆ -k a is s 1 -negative if and only if k ≥ max{1, bh(a) -1}.

Orders on Garside groups

We consider a Garside structure (G, M, ∆) on a Garside group G and two parabolic substructures (H, N, Λ) and (G 1 , M 1 , ∆ 1 ). We assume that

N = M , M 1 = M , N ∪ M 1 generates M , ∆ is central in G, and ∆ 1 is central in G 1 .
Note that the assumption "∆ is central in G" is not so restrictive since, by Dehornoy [5], if (G, M, ∆) is a Garside structure, then (G, M, ∆ k ) is also a Garside structure for each k ≥ 1, and there exists k ≥ 1 such that ∆ k is central in G. We will consider alternating forms with respect to (N, M 1 ).

The depth of an element a ∈ M , denoted by dpt(a), is dpt(a) = bh(a)-1 2 if bh(a) is odd and is dpt(a) = bh(a) 2 if bh(a) is even. In other words, if a = a p • • • a 2 a 1 is the alternating form of a, then dpt(a) is the number of indices i ∈ {1, . . . , p} such that a i ∈ M 1 (that is, the number of even indices). Note that a ∈ M 1 if and only if dpt(a) = 0.

Definition Let α ∈ G and let α = a∆ -k be its ∆-form. We say that α is (H, G 1 )negative if k ≥ 1 and dpt(a) < dpt(∆ k ). We say that α is (H, G 1 )-positive if α -1 is (H, G 1 )-negative. We denote by P = P H,G 1 the set (H, G 1 )-positive elements and by P -1 the set of (H, G 1 )-negative elements.

Definition We say that (H, G 1 ) is a Dehornoy structure if P satisfies the following conditions:

(a) PP ⊂ P,

(b) G 1 PG 1 ⊂ P, (c) we have the disjoint union G = P ⊔ P -1 ⊔ G 1 .
Our goal in this section is to prove a criterion for (H, G 1 ) to be a Dehornoy structure. But, before, we show how the orders appear in this context. Suppose given two sequences of parabolic subgroups G 0 = G, G 1 , . . . , G n and H 1 , . . . ,

H n such that G i+1 , H i+1 ⊂ G i and (H i+1 , G i+1
) is a Dehornoy structure on G i for all i ∈ {0, 1, . . . , n -1} and G n ≃ Z. For each i ∈ {0, 1, . . . , n -1} we denote by P i the set of (H i+1 , G i+1 )-positive elements of G i . On the other hand, we choose a generator α n of G n and we set

P n = {α k n | k ≥ 1}. For each ǫ = (ǫ 0 , ǫ 1 , . . . , ǫ n ) ∈ {±1} n+1 we set P ǫ = P ǫ 0 0 ⊔ P ǫ 1 1 ⊔ • • • ⊔ P ǫn n .
Proposition 3.1 Under the above assumptions P ǫ is the positive cone for a left-order on G.

Proof We must prove that we have a disjoint union G = P ǫ ⊔ (P ǫ ) -1 ⊔ {1} and that P ǫ P ǫ ⊂ P ǫ . The fact that we have a disjoint union

G = P ǫ ⊔ (P ǫ ) -1 ⊔
A with constant ζ if dpt(∆ k ) = ζk + 1 for all k ≥ 1.
We set θ = ∆∆ -1 1 = ∆ -1 1 ∆ ∈ M . We say that an element a ∈ M is a theta element if it is of the form a = θ k a 0 with k ≥ 1 and a 0 ∈ M 1 . We denote by Θ the set of theta elements of M and we set Θ = Θ ∪ M 1 .

Definition Let ζ ≥ 1 be an integer. Let (a, b) ∈ (M × M) \ ( Θ × Θ) such
that a, b are both unmovable. Let ab = c∆ t be the ∆-form of ab. We say that (a, b) satisfies

Condition B with constant ζ if there exists ε ∈ {0, 1} such that (a) dpt(c) = dpt(a) + dpt(b) -ζt -ε, (b) ε = 1 if either a ∈ Θ, or b ∈ Θ, or c ∈ M 1 .
We say that (H, G 1 ) satisfies Condition B with constant Let ζ ≥ 1 be an integer. From here until the end of the section we assume that (H, G 1 ) satisfies Condition A with constant ζ and Condition B with constant ζ . Our goal is then to prove that (H, G 1 ) is a Dehornoy structure, that is, to prove Theorem 3.2.

ζ if each pair (a, b) ∈ (M × M) \ ( Θ × Θ)
Let a be an unmovable element of M and let p = lg(a). Then p is the smallest integer ≥ 0 such that a ≤ R ∆ p . Let com(a) ∈ M such that a com(a) = ∆ p . Then, by El-Rifai-Morton [START_REF] El-Rifai | Algorithms for positive braids[END_REF], com(a) is unmovable, lg(com(a)) = p, and a -1 = com(a)∆ -p is the ∆-form of a -1 . Note that a com(a) = com(a) a = ∆ p since ∆ is central. In particular, com(com(a)) = a.

Lemma 3.3 (1) Let a ∈ M 1 . Then θ ∧ R a = 1 and θ ∨ R a = θa = aθ .
(2) Let a = θ k a 0 be a theta element, where k ≥ 1 and

a 0 ∈ M 1 . Then dpt(a) = ζk + 1. (3) Let a = θ k a 0 be a theta element, where k ≥ 1 and a 0 ∈ M 1 . Then a is unmovable if and only if a 0 is unmovable in M 1 (that is, if and only if ∆ 1 ≤ R a 0 ). (4) Let a be an unmovable element of M . We have a ∈ Θ if and only if com(a) ∈ Θ. (5) Let α ∈ G 1 \ M 1 . Then α has a ∆-form of the form α = a∆ -k where k ≥ 1 and a = θ k a 0 ∈ Θ with a 0 ∈ M 1 . (6) Let a ∈ Θ and b ∈ M \ Θ. Then ab ∈ M \ Θ and ba ∈ M \ Θ. Proof Part (1): Let a ∈ M 1 . Let u = a ∧ R θ . We have u ≤ R θ , hence u∆ 1 ≤ R θ∆ 1 = ∆,
and therefore u∆ 1 ∈ Div(∆). On the other hand, since u ≤ R a, we have

u ∈ M 1 , hence u∆ 1 ∈ M 1 . So, u∆ 1 ∈ Div(∆) ∩ M 1 = Div(∆ 1 ), thus u = 1. Let v = a ∨ R θ .
Since ∆ and ∆ 1 commute with a, we have θa = aθ . In particular,

v ≤ R aθ . Let x 1 ∈ M such that v = x 1 θ . Then x 1 ≤ R a and, since M 1 is a parabolic submonoid, x 1 ∈ M 1 and there exists x 2 ∈ M 1 such that x 2 x 1 = a. So, a = x 2 x 1 ≤ R v = x 1 θ = θx 1 , hence x 2 ≤ R θ , and therefore, since a ∧ R θ = 1, we have x 2 = 1. Thus x 1 = a and v = aθ = θa. Part (2): It is clear that dpt(a) = dpt(aa 0 ) for all a ∈ M and all a 0 ∈ M 1 . Let a = θ k a 0 be a theta element. Then dpt(a) = dpt(θ k ) = dpt(θ k ∆ k 1 ) = dpt(∆ k ) = ζk + 1. Part (3): Let a = θ k a 0 be a theta element. Suppose that ∆ 1 ≤ R a 0 . Let a 1 ∈ M 1 such that a 0 = a 1 ∆ 1 . Then a = θ k a 1 ∆ 1 = θ k-1 a 1 θ∆ 1 = θ k-1 a 1 ∆, hence ∆ ≤ R a. Now suppose that ∆ ≤ R a. By Part (1) we have τ M 1 (a) = a 0 . Since ∆ ≤ R a, we have ∆ 1 ≤ R a, hence ∆ 1 ≤ R τ M 1 (a) = a 0 .
Part (4): Let a be an unmovable element of M and let p = lg(a). Suppose that

a ∈ M 1 . Let b ∈ M 1 such that ab = ∆ p 1 . Then aθ p b = θ p ab = θ p ∆ p 1 = ∆ p , hence com(a) = θ p b ∈ Θ. Suppose that a = θ k a 0 where k ≥ 1 and a 0 ∈ M 1 . We have a = θ k a 0 ≤ R ∆ p = θ p ∆ p 1 hence, by Part (1), a 0 ≤ R ∆ p 1 and k ≤ p. Let b 0 ∈ M 1 such that a 0 b 0 = ∆ p 1 . Then aθ p-k b 0 = θ k a 0 θ p-k b 0 = θ p a 0 b 0 = θ p ∆ p 1 = ∆ p , hence com(a) = θ p-k b 0 ∈ Θ. So, if a ∈ Θ, then com(a) ∈ Θ. Now, since com(com(a)) = a
for each unmovable element a of M , we have a ∈ Θ if and only if com(a) ∈ Θ.

Part (5): Let α ∈ G 1 \ M 1 . Since α ∈ M 1 the ∆ 1 -form of α is of the form α = a∆ -k 1 with a ∈ M 1 , ∆ 1 ≤ R a and k ≥ 1. Then α = a(θ∆ -1 ) k = θ k a∆ -k and θ k a is unmovable by Part (3) of the lemma.
Part (6): Take a, b ∈ M . We assume that a, ab ∈ Θ and we turn to prove that b ∈ Θ. We write ab = θ t c where t ≥ 0 and c ∈ M 1 . On the other hand we know by Part (4) that com(a) ∈ Θ, hence com(a) is of the form com(a) = θ k a 0 with k ≥ 0 and

a 0 ∈ M 1 , and therefore a -1 is of the form a -1 = θ k a 0 ∆ -ℓ = θ k-ℓ a 0 ∆ -ℓ 1 where ℓ = lg(a). So, b∆ ℓ 1 = θ t+k-ℓ a 0 c. If we had t + k -ℓ < 0, then we would have θ ℓ-t-k b∆ ℓ 1 = a 0 c ∈ M 1 , hence we would have θ ℓ-t-k ∈ M 1 , which contradicts Part (1). So, t + k -ℓ ≥ 0. By Part (1) we have τ M 1 (θ t+k-ℓ a 0 c) = a 0 c, hence ∆ 1 ≤ R a 0 c. Let b 0 ∈ M 1 such that b 0 ∆ ℓ 1 = a 0 c. Then b = θ t+k-ℓ b 0 ∈ Θ.
We show in the same way that, if a, ba ∈ Θ, then b ∈ Θ. Lemma 3.4 We have P -1 P -1 ⊂ P -1 .

Proof Let α, β ∈ P -1 . Let α = a∆ -k and β = b∆ -ℓ be the ∆-forms of α and β , respectively. Since α, β ∈ P

-1 , we have k, ℓ ≥ 1, dpt(a) ≤ dpt(∆ k ) -1 = ζk and dpt(b) ≤ dpt(∆ ℓ ) -1 = ζℓ. Let ab = c∆ t be the ∆-form of ab. Then the ∆-form of αβ is αβ = c∆ -k-ℓ+t . We must show that αβ ∈ P -1 , that is, k + ℓ -t ≥ 1 and dpt(c) ≤ dpt(∆ k+ℓ-t ) -1 = ζ(k + ℓ -t). Case 1: a, b ∈ M 1 . Then t = 0 and c = ab, hence k + ℓ -t = k + ℓ ≥ 1 and dpt(c) = 0 ≤ ζ(k + ℓ) = ζ(k + ℓ -t).
) = ζu + 1 ≤ ζℓ, hence u < ℓ. Let ab 0 = c 0 ∆ v 1 be the ∆ 1 -form of ab 0 . If v < u, then t = v and c = θ u-v c 0 , hence k + ℓ -t ≥ ℓ -v ≥ ℓ -u ≥ 1 and dpt(c) = ζ(u -v) + 1 = ζu -ζt + 1 ≤ ζℓ -ζt ≤ ζ(k + ℓ -t). If v ≥ u, then t = u and c = ∆ v-u 1 c 0 ∈ M 1 , hence k + ℓ -t = k + ℓ -u ≥ ℓ -u ≥ 1 and dpt(c) = 0 ≤ ζ(k + ℓ -t).
The case "a ∈ Θ and b ∈ M 1 " can be proved in a similar way.

Case 3: a, b ∈ Θ. We set a = θ u a 0 and b = θ v b 0 , where u, v ≥ 1 and a 0 , b 0 ∈ M 1 . Since dpt(a) = ζu + 1 ≤ ζk, we have u < k. Similarly, we have v < ℓ. Let a 0 b 0 = c 0 ∆ w 1 be the ∆ 1 -form of a 0 b 0 . If w < u + v, then t = w and c = θ u+v-w c 0 , hence k+ℓ-t ≥ k+ℓ-(u+v) = (k-u)+(ℓ-v) ≥ 1 and dpt(c) = ζ(u+v-w)+1 = ζu + 1 + ζv -ζt ≤ ζk + ζℓ -ζt = ζ(k + ℓ -t). If w ≥ u + v, then t = u + v and c = c 0 ∆ w-u-v 1 ∈ M 1 , hence k + ℓ -t = k + ℓ -(u + v) = (k -u) + (ℓ -v) ≥ 1 and dpt(c) = 0 ≤ ζ(k + ℓ -t).
= dpt(c) = dpt(a) + dpt(b) -ζt -1 ≤ ζk + ζℓ -ζt -1 < ζ(k + ℓ -t) .

This (strict) inequality also implies that

k + ℓ -t ≥ 1. If c ∈ M 1 , then 1 ≤ dpt(c) ≤ dpt(a) + dpt(b) -ζt ≤ ζk + ζℓ -ζt = ζ(k + ℓ -t) .
Again, this inequality also implies that k + ℓt ≥ 1.

Lemma 3.5 We have G 1 P -1 G 1 ⊂ P -1 .
Proof We take α ∈ G 1 and β ∈ P -1 and we turn to prove that αβ ∈ P -1 . The proof of the inclusion βα ∈ P -1 is made in a similar way. Let α = a∆ -k and β = b∆ -ℓ be the ∆-forms of α and β , respectively. Since β ∈ P -1 we have ℓ ≥ 1 and dpt(b) ≤ dpt(∆ ℓ )-1 = ζℓ. Let ab = c∆ t be the ∆-form of ab. Then the ∆-form of αβ is αβ = c∆ -k-ℓ+t . We must show that k + ℓt ≥ 1 and dpt(c

) ≤ ζ(k + ℓ -t). Case 1: α ∈ M 1 and b ∈ M 1 . We have k = 0, α = a, t = 0 and c = ab ∈ M 1 . Thus k + ℓ -t = ℓ ≥ 1 and 0 = dpt(c) ≤ ζ(k + ℓ -t).
0 ∈ M 1 . We also have dpt(b) = ζv+1 ≤ ζℓ, hence v < ℓ. Let ab 0 = c 0 ∆ u 1 be the ∆ 1 - form of ab 0 . If u < v, then t = u and c = θ v-u c 0 , hence k + ℓ -t = ℓ -u ≥ ℓ -v ≥ 1 and dpt(c) = ζ(v -u) + 1 = ζv + 1 -ζt ≤ ζℓ -ζt = ζ(k + ℓ -t). If u ≥ v, then t = v and c = ∆ u-v 1 c 0 ∈ M 1 , hence k + ℓ -t = ℓ -v ≥ 1 and 0 = dpt(c) ≤ ζ(k + ℓ -t).
(c) = dpt(a) + dpt(b) -ζt -ε. So, 1 ≤ dpt(c) ≤ 0 + ζℓ -ζt = ζ(k + ℓ -t) .
This inequality also implies that k + ℓt ≥ 1.

Case 4: α ∈ M 1 and b ∈ M 1 . By Lemma 3.3 (5) we have k ≥ 1 and a = θ k a 0 with a 0 ∈ M 1 . Let a 0 b = c 0 ∆ u 1 be the ∆ 1 -form of a 0 b. If u < k, then t = u and c = θ k-u c 0 , hence k + ℓ -t ≥ ℓ ≥ 1 and dpt(c) = ζ(k -u) + 1 ≤ ζk -ζt + ζℓ ≤ ζ(k + ℓ -t). If u ≥ k, then t = k and c = c 0 ∆ u-k 1 ∈ M 1 , hence k + ℓ -t = ℓ ≥ 1 and 0 = dpt(c) ≤ ζ(k + ℓ -t).
+ ℓ -t ≥ k + v -w ≥ 1 and dpt(c) = ζ(k + v -w) + 1 = ζk + ζv + 1 -ζt ≤ ζk + ζℓ -ζt = ζ(k + ℓ -t). If w ≥ k + v, then t = k + v and c = c 0 ∆ w-k-v 1 ∈ M 1 , hence k + ℓ -t = ℓ -v ≥ 1 and 0 = dpt(c) ≤ ζ(k + ℓ -t).
∈ Θ, dpt(c) = dpt(a) + dpt(b) -ζt -1. So, 1 ≤ dpt(c) ≤ ζk + 1 + ζℓ -ζt -1 = ζ(k + ℓ -t) .
This inequality also implies that k + ℓt ≥ 1. Lemma 3. [START_REF] Dehornoy | Alternating normal forms for braids and locally Garside monoids[END_REF] We have G 1 ∩ (P ∪ P -1 ) = ∅.

Proof Let α ∈ G 1 and let α = a∆ -k be the ∆-form of α. If α ∈ M 1 , then k = 0 and α = a, thus α ∈ P -1 . If α ∈ M 1 , then, by Lemma 3.3 (5), we have k ≥ 1 and a = θ k a 0 where a 0 ∈ M 1 , hence dpt(a) = ζk + 1 = dpt(∆ k ), and therefore α ∈ P -1 . Since α -1 ∈ G 1 , we also have α -1 ∈ P -1 , hence α ∈ P. Lemma 3. [START_REF] Dehornoy | Foundations of Garside theory[END_REF] We have P ∩ P -1 = ∅.

Proof Let α ∈ P -1 and let α = a∆ -k be its ∆-form. By definition we have k ≥ 1 and dpt(a

) < dpt(∆ k ) = ζk + 1. Let ℓ = lg(a). Then the ∆-form of α -1 is α -1 = com(a)∆ k-ℓ . We are going to show that α -1 ∈ P -1 , that is, either k -ℓ ≥ 0 or dpt(com(a)) ≥ ζ(ℓ -k) + 1. Case 1: a ∈ M 1 . Let b ∈ M 1 such that ab = ∆ ℓ 1 . We have a -1 = b∆ -ℓ 1 = bθ ℓ ∆ -ℓ = θ ℓ b∆ -ℓ , hence com(a) = θ ℓ b, and therefore, dpt(com(a)) = ζℓ + 1 > ζ(ℓ -k) + 1, since k ≥ 1. So, α -1 ∈ P -1 .
Case 2: a ∈ Θ. We write a = θ u a 0 where a 0 ∈ M 1 and u ≥ 1. We have dpt(a) = ζu + 1 ≤ ζk, hence u < k. Let t ≥ 0 be the length of a 0 and let and therefore α -1 ∈ P -1 .

b 0 ∈ M 1 such that a 0 b 0 = ∆ t 1 . We have a -1 0 = b 0 ∆ -t 1 = θ t b 0 ∆ -t , hence a -1 = θ t-u b 0 ∆ -t , and therefore α -1 = θ t-u b 0 ∆ k-t . If u < t, then com(a) = θ t-u b 0 and dpt(com(a)) = ζ(t -u) + 1 > ζ(t -k) + 1, hence α -1 ∈ P -1 . If u ≥ t, then α -1 = θ -u+t b 0 ∆ k-t = b 0 ∆ u-t 1 ∆ k-t-u+t = b 0 ∆ u-t 1 ∆ k-u and k -u ≥ 1, hence α -1 ∈ P -1 .

Lemma 3.8

We have

G = P ∪ P -1 ∪ G 1 .
Proof We take α ∈ G and we assume that α ∈ (P -1 ∪ G 1 ). We are going to show that α ∈ P, that is, α -1 ∈ P -1 . Let α = a∆ k be the ∆-form of α and let ℓ be the length of a. Then the

∆-form of α -1 is com(a)∆ -k-ℓ . Case 1: a ∈ M 1 . Then k ≥ 1 because α ∈ (P -1 ∪ G 1 ). If a = 1, then α -1 = ∆ -k ∈ P -1
. So, we can assume that a = 1, and therefore

ℓ ≥ 1. Let b ∈ M 1 such that ab = ∆ ℓ 1 . We have a -1 = θ ℓ b∆ -ℓ , hence α -1 = θ ℓ b∆ -k-ℓ and com(a) = θ ℓ b. Then k + ℓ ≥ 1 and dpt(com(a)) = ζℓ + 1 ≤ ζℓ + ζk = ζ(ℓ + k), hence α -1 ∈ P -1 .
Case 2: a ∈ Θ. We write a = θ u a 0 where u ≥ 1 and a 0 ∈ M 1 . Since α ∈ P -1 we have dpt(a

) = ζu + 1 ≥ ζ(-k) + 1, hence u ≥ -k. We also have u = -k, otherwise we would have α = a 0 ∆ -u 1 ∈ G 1 . So, u > -k. Let t be the length of a 0 and let b 0 ∈ M 1 such that a 0 b 0 = ∆ t 1 . We have a -1 0 = b 0 ∆ -t 1 , hence a -1 = θ t-u b 0 ∆ -t , and therefore α -1 = θ t-u b 0 ∆ -k-t . If u < t, then com(a) = θ t-u b 0 , k + t > k + u ≥ 1 and dpt(com(a)) = ζ(t -u) + 1 < ζ(t + k) + 1 = dpt(∆ t+k ), hence α -1 ∈ P -1 . If u ≥ t, then α -1 = b 0 ∆ u-t 1 ∆ -k-u , com(a) = b 0 ∆ u-t 1 ∈ M 1 , k + u ≥ 1, and dpt(com(a)) = 0 ≤ ζ(k + u), hence α -1 ∈ P -1 .
1 ≤ dpt(com(a)) = ζℓ + 1 -dpt(a) ≤ ζℓ + 1 + ζk -1 = ζ(ℓ + k) .
This inequality also implies that ℓ + k ≥ 1. Thus, α -1 ∈ P -1 .

Proof of Theorem 3.2 We have PP ⊂ P by Lemma 3.4, we have G 1 PG 1 ⊂ P by Lemma 3.5, and we have the disjoint union G = P ⊔ P -1 ⊔ G 1 by Lemma 3.6, Lemma 3.7 and Lemma 3.8.

Artin groups of type A

In this section we assume that G and M are the Artin group and the Artin monoid of type A n , respectively, where n ≥ 2. Recall that G is defined by the presentation G = s 1 , . . . , s n | s i s j s i = s j s i s j for |i -j| = 1, s i s j = s j s i for |i -j| ≥ 2 , and that M is the submonoid of G generated by s 1 , s 2 , . . . , s n . Recall also that G is the braid group B n+1 on n + 1 strands and M is the positive braid monoid B + n+1 . By Brieskorn-Saito [START_REF] Brieskorn | Artin[END_REF] and Deligne [START_REF] Deligne | Les immeubles des groupes de tresses généralisés[END_REF], (G, M, Ω) is a Garside structure, where

Ω = (s 1 • • • s n ) • • • (s 1 s 2 s 3 )(s 1 s 2 )s 1 . The element Ω is not central in G but ∆ = Ω 2 = (s 1 • • • s n ) n+1 is
central and, by Dehornoy [5], (G, M, ∆) is also a Garside structure on G. The latter is the Garside structure that we consider in this section.

We denote by G 1 (resp. M 1 ) the subgroup of G (resp. the submonoid of M ) generated by s 2 , . . . , s n and we set

∆ 1 = (s 2 • • • s n ) n . Then (G 1 , M 1 , ∆ 1 ) is a parabolic substruc- ture of (G, M, ∆) and ∆ 1 is central in G 1 .
On the other hand, we denote by H (resp. N ) the subgroup of G (resp. the submonoid of M ) generated by s 1 , . . . , s n-1 and we set Λ = (s 1 • • • s n-1 ) n . Again, (H, N, Λ) is a parabolic substructure of (G, M, ∆). Observe that M 1 ∪ N generates M . The purpose of this section is to prove the following.

1 ≤ i ≤ n -1 we set G i = s i+1 , . . . , s n , M i = s i+1 , . . . , s n + , ∆ i = (s i+1 • • • s n ) n+1-i
and H i = s i , . . . , s n-1 . By iterating Corollary 4.2 and applying Proposition 3.1 we get the following.

Corollary 4.3 (1) For each 1 ≤ i ≤ n -1 the pair (H i , G i ) is a Dehornoy structure on (G i-1 , M i-1 , ∆ i-1 )
, where (G 0 , M 0 , ∆ 0 ) = (G, M, ∆).

(2) For each 1 ≤ i ≤ n -1 we denote by P i the set of (H i , G i )-positive elements of G i-1 . Furthermore we set

P n = {s k n | k ≥ 1}. For each ǫ = (ǫ 1 , . . . , ǫ n ) ∈ {±1} n the set P ǫ = P ǫ 1 1 ⊔ • • • ⊔ P ǫn
n is the positive cone for a left-order on G.

Before proving Theorem 4.1 we show that the orders on G given in Corollary 4.3 (2) coincide with those obtained using Theorem 1.1. More precisely we prove the following.

Proposition 4. [START_REF] Dehornoy | Braid groups and left distributive operations[END_REF] The set P = P H,G 1 of (H, G 1 )-positive elements is equal to the set of s 1 -positive elements of G = B n+1 .

Proof Let P ′ denote the set of s 1 -positive elements of G. We know by Dehornoy [START_REF] Dehornoy | Braid groups and left distributive operations[END_REF] that we have the disjoint union G = P ′ ⊔ P ′-1 ⊔ G 1 . We also know by Corollary 4.2 that PP ⊂ P, G 1 PG 1 ⊂ P and

G = P ⊔ P -1 ⊔ G 1 . Let α ∈ P ′ . By definition α is written α = α 0 s 1 α 1 • • • s 1 α p where p ≥ 1 and α 0 , α 1 , . . . , α p ∈ G 1 . The ∆-form of s 1 is s 1 = s 1 ∆ 0 ,
hence s 1 does not lie in P -1 . The element s 1 does not lie in G 1 either, hence s 1 lies in P. Since PP ⊂ P and G 1 PG 1 ⊂ P we deduce that α lies in P. So, P ′ ⊂ P and therefore P ′-1 ⊂ P -1 . Since we have disjoint unions G = P ⊔ P -1 ⊔ G 1 and G = P ′ ⊔ P ′-1 ⊔ G 1 we conclude that P = P ′ and P -1 = P ′-1 .

The rest of the section is dedicated to the proof of Theorem 4.1. We recall once for all the expressions of ∆ and θ over the standard generators.

∆ = (s 1 s 2 • • • s n ) n+1 = (s 1 • • • s n-1 s 2 n s n-1 • • • s 1 ) • • • (s n-1 s 2 n s n-1 )s 2 n , θ = s 1 • • • s n-1 s 2 n s n-1 • • • s 1 .
Proposition 4. 5 The pair (H, G 1 ) satisfies Condition A with constant ζ = 1.

Proof Let k ≥ 1. Then, by Dehornoy [START_REF] Dehornoy | Alternating normal forms for braids and locally Garside monoids[END_REF], bh

(∆ k ) = bh(Ω 2k ) = 2k + 2, hence dpt(∆ k ) = k + 1.
It remains to show that (H, G 1 ) satisfies Condition B with constant ζ = 1 (see Proposition 4.12). This is the goal of the rest of the section.

An (N, M 1 )-expression of length p of an element a ∈ M is defined to be an expression of a of the form a = a p • • • a 2 a 1 with a i ∈ N if i is even and a i ∈ M 1 if i is odd. Proof Let a ∈ M\M 1 . It suffices to show that bh(aθ) = bh(a)+2. Let a = a p • • • a 2 a 1 be the alternating form of a. Note that, since a ∈ M 1 , we have p ≥ 2. Note also that, by Lemma 3.3 (1), we have

a 1 θ = θa 1 . Then aθ = a p • • • a 3 a 2 θa 1 = a p • • • a 3 b 4 b 3 b 2 a 1 , where b 4 = a 2 s 1 ∈ N , b 3 = s 2 • • • s n-1 s 2 n ∈ M 1 and b 2 = s n-1 • • • s 2 s 1 ∈ N . We turn to show that aθ = a p • • • a 2 b 4 b 3 b 2 a 1
is the alternating form of aθ . This will prove the lemma.

Let x = τ M 1 (a p • • • a 3 b 4 b 3 b 2 ) = τ M 1 (a p • • • a 3 a 2 θ). We know by Lemma 3.3 (1) that x ∨ R θ = θx = xθ , hence x ≤ R a p • • • a 2 , and therefore x = 1, since τ M 1 (a p • • • a 3 a 2 ) = 1. We have a p • • • a 3 b 4 b 3 = a p • • • a 3 a 2 s 1 s 2 • • • s n-1 s 2 n . It is easily checked that (s 1 • • • s n-1 s 2 n ) ∨ R s i = s i+1 (s 1 • • • s n-1 s 2 n ) for all i ∈ {1, . . . , n -1}. Thus, if there exists i ∈ {1, . . . , n-1} such that s i ≤ R a p • • • a 3 b 4 b 3 , then there exists j ∈ {2, . . . , n} such that s j ≤ R a p • • • a 3 a 2 . But, since τ M 1 (a p • • • a 3 a 2 ) = 1, such a j does not exist, hence such an i does not exist either, hence τ N (a p • • • a 3 b 4 b 3 ) = 1. We have a p • • • a 3 b 4 = a p • • • a 3 a 2 s 1 .
We have s 1 ∨ R s i = s i s 1 for all i ∈ {3, . . . , n}, and

s 1 ∨ R s 2 = s 1 s 2 s 1 . Thus, for i ∈ {2, . . . , n}, if s i ≤ R a p • • • a 3 b 4 , then s i ≤ R a p • • • a 3 a 2 .
Since such an i does not exist, we have

τ M 1 (a p • • • a 3 b 4 ) = 1. This finishes the proof that a p • • • a 3 b 4 b 3 b 2 a 1 is the alternating form of aθ since a p • • • a 3 is an alternating form and τ N (a p • • • a 3 ) = 1. Lemma 4.9 (1) Let a ∈ M 1 and b ∈ M \ M 1 . Then dpt(ab) = dpt(ba) = dpt(b).
(

≥ 0. If a∆ -k ∈ G 1 then a ∈ Θ. Proof Let a∆ -k = a 0 ∆ -t 1 be the ∆ 1 -form of a∆ -k . We have a = a 0 ∆ -t 1 ∆ k = θ k a 0 ∆ k-t 1 . If k ≥ t then we clearly have a ∈ Θ. Suppose that k < t. Then a∆ t-k 1 = θ k a 0 , hence ∆ t-k 1 ≤ R τ M 1 (θ k a 0 ). By Lemma 3.3 (1) we have τ M 1 (θ k a 0 ) = a 0 , hence ∆ t-k 1 ≤ R a 0 . Let b 0 ∈ M 1 such that a 0 = b 0 ∆ t-k 1 . Then a = θ k b 0 ∈ Θ. Lemma 4.11 Let a, b ∈ M \ M 1 , c ∈ M 1 and k ≥ 0 such that ab = c∆ k and dpt(a) + dpt(b) = k + 2. Then (a, b) ∈ (Θ × Θ).
Proof Let p = dpt(a) and q = dpt(b). Note that, since a, b ∈ M 1 , we have p, q ≥ 1.

We have bh(a) ≥ 2p, hence bh(a) -1 > 2p -2, and therefore, by Theorem 2.5, Ω -2p+2 a = a∆ -p+1 either lies in G 1 or is s 1 -positive. Similarly, b∆ -q+1 either lies in G 1 or is s 1 -positive. If either a∆ -p+1 was s 1 -positive or b∆ -q+1 was s 1 -positive, then c = ab∆ -k = (a∆ -p+1 )(b∆ -q+1 ) would be s 1 -positive. Since c ∈ M 1 , c cannot be s 1 -positive, hence both a∆ -p+1 and b∆ -q+1 lie in G Case 3: a, b ∈ M \ Θ. Set p = dpt(a) and q = dpt(b). We have bh(a) ∈ {2p, 2p + 1} hence, by Theorem 2.5, Ω -2p a is s 1 -negative and Ω -2p+2 a either lies in G 1 or is

s 1 -positive. Similarly, Ω -2q b is s 1 -negative and Ω -2q+2 b either lies in G 1 or is s 1 -positive. So, Ω -2p-2q
ab is s 1 -negative and Ω -2p-2q+4 ab either lies in G 1 or is s 1 -positive. By Theorem 2.5 it follows that bh(ab) -1 ≤ 2p + 2q and 2p + 2q -4 < bh(ab) -1, hence 2p + 2q -2 ≤ bh(ab) ≤ 2p + 2q + 1, and therefore p + q -1 ≤ dpt(ab) ≤ p + q. So, there exists ε ∈ {0, 1} such that dpt(ab

) = p + q -ε = dpt(a) + dpt(b) -ε.
Suppose that c ∈ M 1 . By Lemma 4.9 (2), dpt(c)

+ t = dpt(c) + dpt(∆ t ) -1 = dpt(c∆ t ) = dpt(ab) = dpt(a) + dpt(b) -ε, hence dpt(c) = dpt(a) + dpt(b) -t -ε. Suppose that c ∈ M 1 . By Lemma 4.9 (1), dpt(a) + dpt(b) -ε = dpt(ab) = dpt(c∆ t ) = dpt(∆ t ) = t + 1, hence dpt(a) + dpt(b) = t + 1 + ε. Since a, b ∈ Θ Lemma 4.11 implies that ε = 0. So, dpt(c) = 0 = dpt(a) + dpt(b) -t -1.
5 Artin groups of dihedral type, the even case

Let m ≥ 4 be an integer. Recall that the Artin group of type I 2 (m) is the group

G = A I 2 (m) defined by the presentation G = s, t | Π(s, t, m) = Π(t, s, m
) . Let M be the submonoid of G generated by {s, t} and let Ω = Π(s, t, m). Then, by Brieskorn-Saito [START_REF] Brieskorn | Artin[END_REF] and Deligne [START_REF] Deligne | Les immeubles des groupes de tresses généralisés[END_REF], the triple (G, M, Ω) is a Garside structure on G. If m is even then ∆ = Ω is central. However, if m is odd then Ω is not central but ∆ = Ω 2 is central. In both cases, by Dehornoy [5], the triple (G, M, ∆) is a Garside structure on G. In this section we study the case where m is even and in the next one we will study the case where m is odd. So, from now until the end of the section we assume that m = 2k is even and

∆ = Π(s, t, m) = (st) k = (ts) k .
Remark By setting ∆ = Ω 2 in the even case as in the odd case we could state global results valid for all m ≥ 4, but it would be still necessary to differentiate the even case from the odd case in the proofs, and this would lengthen the proofs for the even case.

We denote by G 1 (resp. M 1 ) the subgroup of G (resp. submonoid of M ) generated by t, and by H (resp. N ) the subgroup of G (resp. submonoid of M ) generated by s. We set ∆ 1 = t and Λ = s. By Brieskorn-Saito [START_REF] Brieskorn | Artin[END_REF] the triples (G 1 , M 1 , ∆ 1 ) and (H, N, Λ) are parabolic substructures of (G, M, ∆). On the other hand it is obvious that M 1 ∪ N generates M . The main result of the present section is the following. By Theorem 3.2 this implies the following.

Corollary 5. [START_REF] Burckel | The wellordering on positive braids[END_REF] The pair (H, G 1 ) is a Dehornoy structure on G.

We denote by P 1 the set of (H, G 1 )-positive elements of G and we set

P 2 = {t n | n ≥ 1}. For each ǫ = (ǫ 1 , ǫ 2 ) ∈ {±1} 2 we set P ǫ = P ǫ 1 1 ∪ P ǫ 2 2 .
Then, by Proposition 3.1, we have the following.

Corollary 5. [START_REF] Crisp | Injective maps between Artin groups, Geometric group theory down under[END_REF] The set P ǫ is the positive cone for a left-order on G.

In this section we denote by r 1 , . . . , r 2k-1 the standard generators of the braid group B 2k on 2k = m strands. By Crisp [START_REF] Crisp | Injective maps between Artin groups, Geometric group theory down under[END_REF] we have an embedding ι : G → B 2k which sends s to k-1 i=0 r 2i+1 and sends t to k-1 i=1 r 2i . In the second part of the section we will show that the orders obtained from Corollary 5.3 can be deduced from ι together with the Dehornoy order. More precisely, we show the following. Proposition 5.4 Let α ∈ G. Then α is (H, G 1 )-negative if and only if ι(α) is r 1 -negative.

The proof of Theorem 5.1 is based on the following observation whose proof is left to the reader. Lemma 5.5 Let a be an unmovable element of M . Then a is uniquely written in the form a = t up s vp • • • t u 1 s v 1 t u 0 with u 0 , u p ≥ 0, u 1 , . . . , u p-1 ≥ 1 and v 1 , . . . , v p ≥ 1. In this case dpt(a) = p.

The first part of Theorem 5.1 is a straightforward consequence of this lemma. Proof Let p ≥ 1 be an integer. We have θ = s(ts) k-1 , hence θ p = (s(ts) k-1 ) p . By Lemma 5.5 it follows that dpt(θ

p ) = p(k -1) + 1, hence dpt(∆ p ) = dpt(θ p t p ) = dpt(θ p ) = p(k -1) + 1.
If a ∈ M \ {1} is written as in Lemma 5. ( We turn now to the proof of Proposition 5.4. We denote by G ′ 1 (resp M ′ 1 ) the subgroup of B 2k (resp. the submonoid of B + 2k ) generated by r 2 , . . . , r 2k-1 and we denote by H ′ (resp. N ′ ) the subgroup of B 2k (resp. the submonoid of B + 2k ) genetared by r 1 , . . . , r 2k-2 . Note that ι(t) ∈ G ′ 1 , hence ι(G 1 ) ⊂ G ′ 1 . We denote by

Ω B = (r 1 r 2 • • • r 2k-1 ) • • • (r 1 r 2 )r 1 the standard Garside element of B 2k and by Φ : B 2k → B 2k , α → Ω B αΩ -1 B , the conjugation by Ω B . Recall that Φ(r i ) = r 2k-i for all i ∈ {1, . . . , 2k -1}. So, Φ(G ′ 1 ) = H ′ and Φ(H ′ ) = G ′ 1 .
Lemma 5.9 Let a be an unmovable element of M such that dpt(a) ≤ k -1. Then there exist b

1 ∈ M ′ 1 and b 2 ∈ N ′ such that ι(a) = b 1 b 2 .
Proof Let p = dpt(a). By Lemma 5.5, a can be written a = t u 0 s v 1 t u 1 • • • s vp t up where u 0 , u p ≥ 0, u 1 , . . . , u p-1 ≥ 1 and v 1 , . . . , v p ≥ 1. We show by induction on p that there exist b 1 ∈ M ′ 1 and b 2 ∈ r 1 , . . . , r 2p + such that ι(a) = b 1 b 2 . Since p ≤ k -1 this proves the lemma. The case p = 0 is obvious because ι(t) ∈ M ′ 1 . We assume that 1 ≤ p ≤ k -1 and that the inductive hypothesis holds. Set

a ′ = t u 0 s v 1 t u 1 • • • s v p-1 t u p-1 . By induction there exist b ′ 1 ∈ M ′ 1 and b ′ 2 ∈ r 1 , . . . , r 2p-2 + such that ι(a ′ ) = b ′ 1 b ′ 2 . Note that b ′ 2 commutes with r i for all i ≥ 2p. So, ι(a) = b ′ 1 b ′ 2 k-1 i=0 r vp 2i+1 k-1 i=1 r up 2i = b ′ 1 k-1 i=p r vp 2i+1 b ′ 2 p-1 i=0 r vp 2i+1 k-1 i=1 r up 2i = b ′ 1 k-1 i=p r vp 2i+1   k-1 i=p+1 r up 2i   b ′ 2 p-1 i=0 r vp 2i+1 p i=1 r up 2i = b 1 b 2 ,
where

b 1 = b ′ 1 k-1 i=p r vp 2i+1   k-1 i=p+1 r up 2i   ∈ M ′ 1 , b 2 = b ′ 2 p-1 i=0 r vp 2i+1 p i=1 r up 2i
∈ r 1 , . . . , r 2p + .

Proof of Proposition 5. [START_REF] Dehornoy | Braid groups and left distributive operations[END_REF] We denote by P the set of (H, G 1 )-positive elements of G and by P ′ the set of r 1 -positive elements of B 2k . By Corollary 5.2 we have the disjoint union G = P ⊔ P -1 ⊔ G 1 and by Dehornoy [START_REF] Dehornoy | Braid groups and left distributive operations[END_REF] we have the disjoint union

B 2k = P ′ ⊔ P ′-1 ⊔ G ′ 1 .
It suffices to show that ι(P -1 ) ⊂ P ′-1 . Indeed, suppose that ι(P -1 ) ⊂ P ′-1 . Since ι is a homomorphism we also have ι(P) ⊂ P ′ . Since we also know that ι(G 1 ) ⊂ G ′ 1 , from the disjoint unions given above follows that α ∈ P -1 if and only if ι(α) ∈ P ′-1 .

Let α be an element of P -1 . Let α = a∆ -p be the ∆-form of α. By definition we have p ≥ 1 and dpt(a) ≤ p(k -1). Suppose first that p = 1 and dpt(a) ≤ k -1. By Lemma 5.9 there exist b

1 ∈ M ′ 1 and b 2 ∈ N ′ such that ι(a) = b 1 b 2 . Moreover, by Crisp [3], ι(∆) = Ω B . Thus ι(α) = b 1 b 2 Ω -1 B = b 1 Ω -1 B Φ(b 2 ). Since b 1 , Φ(b 2 ) ∈ M ′ 1
and Ω -1 B ∈ P ′-1 , it follows that ι(α) ∈ P ′-1 . Now we consider the general case where p ≥ 1 and dpt(a) ≤ p(k -1). It is easily deduced from Lemma 5.5 that a can be written a = a 1 a 2 • • • a p where a i is an unmovable element of M such that dpt(a i ) ≤ k -1 for all i ∈ {1, . . . , p}. Note that a i may be equal to 1 in the above expression. We have α = (a 1 ∆ -1 )(a 2 ∆ -1 ) • • • (a p ∆ -1 ) and, by the above, ι(a i ∆ -1 ) ∈ P ′-1 for all i ∈ {1, . . . , p}, hence ι(α) ∈ P ′-1 .

Artin groups of dihedral type, the odd case

Let m = 2k + 1 ≥ 5 be an odd integer and let G = A I 2 (m) = s, t | Π(s, t, m) = Π(t, s, m) be the Artin group of type I 2 (m). Let M be the submonoid of G generated by {s, t} and let Ω = Π(s, t, m) = (st) k s = (ts) k t. Recall that, by Brieskorn-Saito [START_REF] Brieskorn | Artin[END_REF] and Deligne [START_REF] Deligne | Les immeubles des groupes de tresses généralisés[END_REF], the triple (G, M, Ω) is a Garside structure on G. As pointed out in Section 5, Ω is not central but ∆ = Ω 2 is, and, by Dehornoy [5], (G, M, ∆) is also a Garside structure on G. This is the Garside structure on G that will be considered in the present section.

We denote by G 1 (resp. M 1 ) the subgroup of G (resp. submonoid of M ) generated by t, and by H (resp. N ) the subgroup of G (resp. submonoid of M ) generated by s. Set ∆ 1 = t 2 and Λ = s 2 . Then, by Brieskorn-Saito [START_REF] Brieskorn | Artin[END_REF], the triples (G 1 , M 1 , ∆ 1 ) and (H, N, Λ) are parabolic substructures of (G, M, ∆). Moreover, M 1 ∪ N obviously generates M . The main result of this section is the following. (

) Let a, b ∈ M \ {1} such that ab = Ω. Then dpt(a) + dpt(b) = k + 1 if σ(a) = s , k if σ(a) = t . 2
(3) Let c be an Ω-unmovable element of M . Then

dpt(ϕ(c)) =    dpt(c) + 1 if c = 1 and σ(c) = τ (c) = t , dpt(c) -1 if c = 1 and σ(c) = τ (c) = s , dpt (c) otherwise .
(4) Let c be an Ω-unmovable element of M . Then Lemma 6.9 Let a, b be two Ω-unmovable elements in M . We assume that the ∆-form of ab is in the form ab = c∆ p where c is Ω-unmovable.

dpt(cΩ) = dpt(c) + k -1 if c = 1 and τ (c) = s , dpt(c) + k otherwise . (5) Let a, b ∈ M \ {1} such that a ϕ(b) = Ω. Then dpt(a) + dpt(b) = k + 1 if τ (a) = s , k if τ (a) = t . Lemma 6.8 Let a 1 , a 2 , b 1 , b 2 be four non-trivial Ω-unmovable elements of M such that σ(a 1 ) = τ (a 2 ), τ (b 1 ) = σ(b 2 ), a 1 b 1 = Ω and a 2 ϕ(b 2 ) = Ω. Set u = |{i ∈ {1, 2} | σ(a i ) = s}| and v = |{i ∈ {1, 2} | τ (b i ) = s}|. Then dpt(a 1 ) + dpt(a 2 ) + dpt(b 1 ) + dpt(b 2 ) = 2k -1 + u + v. Proof If σ(a 1 ) = s and σ(a 2 ) = s, then τ (a 2 ) = s, τ (b 1 ) = s and τ (b 2 ) = t, hence u = 2, v = 1 and, by Lemma 6.7, dpt(a 1 ) + dpt(a 2 ) + dpt(b 1 ) + dpt(b 2 ) = 2k + 2 = 2k -1 + u + v. If σ(a 1 ) = s and σ(a 2 ) = t, then τ (a 2 ) = s, τ (b 1 ) = s and τ (b 2 ) = s, hence u = 1, v = 2
(1) Suppose that (a, b) ∈ ( Θ × Θ). There exists ε ∈ {0, 1} such that dpt

(c) = dpt(a) + dpt(b) -p(2k -1) -ε. Moreover, ε = 1 if either a ∈ Θ or b ∈ Θ or c ∈ M 1 .
( (3) Suppose that (a, bΩ) ∈ ( Θ × Θ). There exists ε ∈ {0, 1} such that dpt(cΩ) = dpt(a) + dpt(bΩ)p(2k -1)ε. Moreover, ε = 1 if either a ∈ Θ or bΩ ∈ Θ.

(4) Suppose that (aΩ, ϕ(b) Ω) ∈ ( Θ × Θ). There exists ε ∈ {0, 1} such that • a i = 1, b i = 1, a i b i = Ω if i is odd, and a i ϕ(b i ) = Ω if i is even, for all i ∈ {1, . . . , 2p};

• c = a 2p+1 b 2p+1 ;

• Set x i = τ (a i ), x ′ i = σ(a i ), y i = σ(b i ), y ′ i = τ (b i ), for all i ∈ {1, . . . , 2p + 1}. Then x i+1 = x ′ i for all i ∈ {1, . . . , 2p -1}. We have y i = ϕ(x i ) and y ′ i = x ′ i if i is odd, and y i = x i and y ′ i = ϕ(x ′ i ) if i is even, for all i ∈ {1, . . . , 2p}. Thus, if i is odd, then y i+1 = x i+1 = x ′ i = y ′ i , and if i is even, then y i+1 = ϕ(x i+1 ) = ϕ(x ′ i ) = y ′ i , for i ∈ {1, . . . , 2p -1}. Let u = |{i ∈ {1, . . . , 2p} | x ′ i = s}|. By using Lemma 6.7 we show successively the following equalities. dpt(a) = dpt(a 2p+1 ) + 2p i=1 dpt(a i )u + ε 1,a , dpt(aΩ) = dpt(a 2p+1 ) + 2p i=1 dpt(a i )u + k + ε 2,a , where ε 1,a and ε 2,a are as follows. If p ≥ 1 and a 2p+1 = 1, then: If p ≥ 1 and a 2p+1 = 1, then:

ε 1,a = 0 if (x ′ 2p , x 2p+1 ) ∈ {(s,
ε 1,a = 0 if x ′ 2p = t , 1 if x ′ 2p = s , ε 2,a =      -1 if (x 1 , x ′ 2p ) = (s, t) , 0 if (x 1 , x ′ 2p ) ∈ {(s, s), (t, t)} , 1 if (x 1 , x ′ 2p ) = (t, s) .
If p = 0 and a = 1, then:

ε 1,a = 0 , ε 2,a = -1 if x 2p+1 = s , 0 if x 2p+1 = t .
If p = 0 and a = 1, then ε 1,a = ε 2,a = 0.

Let v = |{i ∈ {1, . . . , 2p} | y ′ i = s}|. Similarly, by using Lemma 6.7 we prove successively the following equalities. IMB UMR 5584, CNRS, Univ. Bourgogne Franche-Comté, 21000 Dijon, France IMB, UMR 5584,CNRS, Univ. Bourgogne Franche-Comté, 21000 Dijon, France arcisd@gmail.com, lparis@u-bourgogne.fr

  as above satisfies Condition B with constant ζ . Theorem 3.2 If there exists a constant ζ ≥ 1 such that (H, G 1 ) satisfies Condition A with constant ζ and Condition B with constant ζ , then (H, G 1 ) is a Dehornoy structure.

Case 2 :

 2 a ∈ M 1 and b ∈ Θ. We write b = θ u b 0 where u ≥ 1 and b 0 ∈ M 1 . By Lemma 3.3 (3) we have dpt(b

Case 4 :

 4 either a ∈ Θ, or b ∈ Θ. Since (H, G 1 ) satisfies Condition B with constant ζ , there exists ε ∈ {0, 1} such that dpt(c) = dpt(a) + dpt(b)ζtε. If c ∈ M 1 , then ε = 1 and 0

Case 2 :

 2 α ∈ M 1 and b ∈ Θ. We have k = 0, α = a and b = θ v b 0 where v ≥ 1 and b

  Case 3: α ∈ M 1 and b ∈ M \ Θ. We have k = 0 and α = a. On the other hand, by Lemma 3.3 (6), we have ab ∈ M \ Θ, hence c ∈ M 1 , and therefore dpt(c) ≥ 1. Since (H, G 1 ) satisfies Condition B with constant ζ , there exists ε ∈ {0, 1} such that dpt

Case 5 :

 5 α ∈ M 1 and b ∈ Θ. By Lemma 3.3 (5) we have k ≥ 1 and a = θ k a 0 with a 0 ∈ M 1 . On the other hand, b is written b = θ v b 0 with v ≥ 1 and b 0 ∈ M 1 . Since dpt(b) = ζv + 1 ≤ ζℓ, we have v < ℓ. Let a 0 b 0 = c 0 ∆ w 1 be the ∆ 1 -form of a 0 b 0 . If w < k + v, then t = w and c = θ k+v-w c 0 , hence k

Case 6 :

 6 α ∈ M 1 and b ∈ M \ Θ. By Lemma 3.3 (5) we have k ≥ 1 and a = θ k a 0 with a 0 ∈ M 1 . On the other hand, by Lemma 3.3 (6), we have ab ∈ M \ Θ, hence c ∈ M 1 , and therefore dpt(c) ≥ 1. Since (H, G 1 ) satisfies Condition B with constant ζ and a

Case 3 :

 3 a ∈ M \ Θ. Recall that a com(a) = ∆ ℓ . Since (H, G 1 ) satisfies Condition B with constant ζ and 1 ∈ M 1 , we have 0 = dpt(1) = dpt(a) + dpt(com(a))ζℓ -1, hence dpt(com(a)) = ζℓ + 1 -dpt(a) ≥ ζℓ + 1ζk = ζ(ℓk) + 1 ,

Case 3 :

 3 a ∈ M \ Θ. Since (H, G 1 ) satisfies Condition B with constant ζ , we have 0 = dpt(1) = dpt(a) + dpt(com(a))ζℓ -1. On the other hand, since ∆ ℓ ∈ Θ, by Lemma 3.3 (6), com(a) ∈ Θ, hence com(a) ∈ M 1 , and therefore dpt(com(a)) ≥ 1. Moreover, since α ∈ P -1 , we have dpt(a) ≥ ζ(-k) + 1. So,

Theorem 4 . 1 Corollary 4 . 2

 4142 The pair (H, G 1 ) satisfies Condition A with constant ζ = 1 and Condition B with constant ζ = 1.By applying Theorem 3.2 we deduce the following. The pair (H, G 1 ) is a Dehornoy structure.

For

Lemma 4 . 6 (Lemma 4 . 7

 4647 Dehornoy [6],Burckel [2]) Let a ∈ M and let a = a p • • • a 2 a 1 be an (N, M 1 )-expression of a. Then p ≥ bh(a).Let a ∈ M . Choose an expression a = s i ℓ • • • s i 2 s i 1 of a over S and set rev(a) = s i 1 s i 2 • • • s i ℓ .Since the relations that define M are symmetric, the definition of rev(a) does not depend on the choice of the expression of a. It is easily checked that rev(Ω) = Ω, rev(∆) = ∆ and rev(θ) = θ . Moreover, rev(a) ∈ M 1 for all a ∈ M 1 and rev(a) ∈ N for all a ∈ N . Let a ∈ M . Then dpt(rev(a)) = dpt(a). Proof Let a = a p • • • a 2 a 1 be the alternating form of a. If p is even, then rev(a) = rev(a 1 ) rev(a 2 ) • • • rev(a p ) 1 is a (N, M 1 )-expression of rev(a) hence, by Lemma 4.6, p + 1 ≥ bh(rev(a)), and therefore dpt(a) = p 2 ≥ dpt(rev(a)). If p is odd, then rev(a) = rev(a 1 ) rev(a 2 ) • • • rev(a p ) is a (N, M 1 )-expression of rev(a) hence, by Lemma 4.6, p ≥ bh(rev(a)), and therefore dpt(a) = p-1 2 ≥ dpt(rev(a)). So, dpt(a) ≥ dpt(rev(a)) in both cases. Since rev(rev(a)) = a, we also have dpt(rev(a)) ≥ dpt(a), hence dpt(rev(a)) = dpt(a).

Lemma 4 . 8

 48 Let a ∈ M \ M 1 and k ≥ 1. Then dpt(aθ k ) = dpt(a) + k.

  ) Let a ∈ Θ and b ∈ M \ M 1 . Then dpt(ab) = dpt(ba) = dpt(a) + dpt(b) -1. Proof Let a ∈ M 1 and b ∈ M \ M 1 . We obviously have bh(ba) = bh(b), hence dpt(ba) = dpt(b). On the other hand, since rev(a) ∈ M 1 , By Lemma 4.7 we have dpt(ab) = dpt(rev(ab)) = dpt(rev(b) rev(a)) = dpt(rev(b)) = dpt(b). Let a ∈ Θ and b ∈ M \ M 1 . Write a = θ k a 0 with a 0 ∈ M 1 and k ≥ 1. By the above and Proposition 4.5 we have dpt(a) = dpt(θ k ) = dpt(∆ k ) = k + 1. Then, by the above and Lemma 4.8, dpt(ba) = dpt(bθ k ) = dpt(b) + k = dpt(a) + dpt(b) -1. On the other hand, since rev(a) ∈ Θ, we have dpt(ab) = dpt(rev(ab)) = dpt(rev(b) rev(a)) = dpt(rev(a)) + dpt(rev(b)) -1 = dpt(a) + dpt(b) -1.Lemma 4.10 Let a ∈ M and k

1 .

 1 We conclude by Lemma 4.10 that a, b ∈ Θ, hence a, b ∈ Θ since we assumed that a, b ∈ M 1 . Now we are ready to prove the second part of Theorem 4.1. Proposition 4.12 The pair (H, G 1 ) satisfies Condition B with constant ζ = 1. Proof We take (a, b) ∈ (M × M) \ ( Θ × Θ) such that a and b are unmovable. We must show that (a, b) satisfies Condition B with constant ζ = 1. Let ab = c∆ t be the ∆-form of ab. So, we must show that there exists ε ∈ {0, 1} such that dpt(c) = dpt(a) + dpt(b)tε, and ε = 1 if either a ∈ Θ, or b ∈ Θ, or c ∈ M 1 . Case 1: a ∈ M 1 and b ∈ M \ Θ. By Lemma 3.3 (6) we have ab ∈ Θ, hence c ∈ M 1 . Then, by Lemma 4.9, dpt(a) + dpt(b) = dpt(b) = dpt(ab) = dpt(c) + t, hence dpt(c) = dpt(a) + dpt(b)t -0. The case a ∈ M \ Θ and b ∈ M 1 is proved in a similar way. Case 2: a ∈ Θ and b ∈ M \ Θ. We write a = θ k a 0 where k ≥ 1 and a 0 ∈ M 1 . Again, by Lemma 3.3 (6) we have ab ∈ Θ, hence c ∈ M 1 . Then, by Lemma 4.9, dpt(a) + dpt(b) -1 = dpt(ab) = dpt(c) + t, hence dpt(c) = dpt(a) + dpt(b)t -1. The case a ∈ M \ Θ and b ∈ Θ is proved in a similar way.

Theorem 5 . 1

 51 The pair (H, G 1 ) satisfies Condition A with constant ζ = k -1 and Condition B with constant ζ = k -1.

Proposition 5 . 6

 56 The pair (H, G 1 ) satisfies Condition A with constant ζ = k -1.

  5 we set σ(a) = t if u p = 0 and σ(a) = s if u p = 0. Similarly we set τ (a) = t if u 0 = 0 and τ (a) = s if u 0 = 0. In other words σ(a) is the first letter of a and τ (a) is the last one. The following is a straightforward consequence of Lemma 5.5. Lemma 5.7 (1) Let a, b be two unmovable elements of M such that ab is unmovable. Then dpt(ab) = dpt(a) + dpt(b) -1 if a = 1, b = 1 and τ (a) = σ(b) = s , dpt(a) + dpt(b) otherwise .

1 . 5 . 8

 158 ) Let a, b ∈ M such that ab = ∆. Then dpt(a) + dpt(b) = dpt(∆) = k. Now we can prove the second part of Theorem 5.Proposition The pair (H, G 1 ) satisfies Condition B with constant ζ = k -1. Proof We take two unmovable elements a, b ∈ M such that (a, b) ∈ Θ × Θ and we denote by ab = c∆ p the ∆-form of ab. We must show that there exists ε ∈ {0, 1} such that dpt(c) = dpt(a) + dpt(b)p(k -1)ε and that ε = 1 if either a ∈ Θ, or b ∈ Θ, or c ∈ M 1 . We write a = a p+1 a p • • • a 1 and b = b 1 • • • b p b p+1 so that: • a i = 1, b i = 1 and a i b i = ∆ for all i ∈ {1, . . . , p}; a p+1 = 1, b p+1 = t r-q b ′ and (x ′ p , y p+1 ) = (s, t), hence ε = 1. The case b ∈ Θ is proved in the same way. Suppose that c ∈ M 1 . Then p ≥ 1, since (a, b) ∈ ( Θ × Θ). If a p+1 = 1 and b p+1 = 1, then (x p+1 , y p+1 ) = (t, t), hence ε = 1. If a p+1 = 1 and b p+1 = 1, then x p+1 = t, hence ε = 1. If a p+1 = 1 and b p+1 = 1, then y p+1 = t, hence ε = 1. If a p+1 = 1 and b p+1 = 1, then ε = 1.

Theorem 6 . 1

 61 The pair (H, G 1 ) satisfies Condition A with constant ζ = 2k -1 and Condition B with constant ζ = 2k -1.

Lemma 6 . 7 ( 1 )

 671 Let a, b ∈ M such that ab is Ω-unmovable. Then dpt(ab) = dpt(a) + dpt(b) -1 if a = 1, b = 1, and τ (a) = σ(b) = s , dpt(a) + dpt(b) otherwise .

  and, by Lemma 6.7, dpt(a 1 ) + dpt(a 2 )+ dpt(b 1 ) + dpt(b 2 ) = 2k + 2 = 2k -1 + u + v. If σ(a 1 ) = t and σ(a 2 ) = s, then τ (a 2 ) = t, τ (b 1 ) = t and τ (b 2 ) = t, hence u = 1, v = 0 and, by Lemma 6.7, dpt(a 1 ) + dpt(a 2 ) + dpt(b 1 ) + dpt(b 2 ) = 2k = 2k -1 + u + v. If σ(a 1 ) = t and σ(a 2 ) = t, then τ (a 2 ) = t, τ (b 1 ) = t and τ (b 2 ) = s, hence u = 0, v =1 and, by Lemma 6.7, dpt(a 1 ) + dpt(a 2 ) + dpt(b 1 ) + dpt(b 2 ) = 2k = 2k -1 + u + v.

)

  Suppose that (aΩ, ϕ(b)) ∈ ( Θ × Θ). The exists ε ∈ {0, 1} such that dpt(cΩ) = dpt(aΩ) + dpt(ϕ(b))p(2k -1)ε. Moreover, ε = 1 if either aΩ ∈ Θ or ϕ(b) ∈ Θ.

  dpt(c) = dpt(aΩ) + dpt(ϕ(b) Ω) -(p + 1)(2k -1)ε. Moreover, ε = 1 if either aΩ ∈ Θ or ϕ(b) Ω ∈ Θ or c ∈ M 1 . Proof We write a and b in the form a = a 2p+1 a 2p • • • a 2 a 1 and b = b 1 b 2 • • • b 2p b 2p+1 so that:

  dpt(b) = dpt(b 2p+1 ) + 2p i=1 dpt(b i )v + ε 1,b , dpt(ϕ(b)) = dpt(b 2p+1 ) + 2p i=1 dpt(b i )v + ε 2,b , dpt(bΩ) = dpt(b 2p+1 ) + 2p i=1 dpt(b i )v + k + ε 3,b , dpt(ϕ(b) Ω) = dpt(b 2p+1 ) + 2p i=1 dpt(b i )v + k + ε 4,b ,where ε 1,b , ε 2,b , ε 3,b and ε 4,b are as follows. If p ≥ 1 and b 2p+1 = 1, then: ε 1,b = 0 if (y

( 2 )( 3 )

 23 Suppose that (aΩ, ϕ(b)) ∈ ( Θ × Θ). There exists ε ∈ {0, 1} such that dpt(c) = dpt(aΩ)+ dpt(ϕ(b)) -(p + 1)(2k -1)ε. Moreover, ε = 1 if either aΩ ∈ Θ or ϕ(b) ∈ Θ or c ∈ M 1 . Suppose that (a, bΩ) ∈ ( Θ × Θ). There exists ε ∈ {0, 1} such that dpt(c) = dpt(a) + dpt(bΩ) -(p + 1)(2k -1)ε. Moreover, ε = 1 if either a ∈ Θ or bΩ ∈ Θ or c ∈ M 1 . (4) Suppose that (aΩ, ϕ(b)Ω) ∈ ( Θ × Θ). There exists ε ∈ {0, 1} such that dpt(cΩ) = dpt(aΩ) + dpt(ϕ(b) Ω) -(p + 1)(2k -1)ε. Moreover, ε = 1 if either aΩ ∈ Θ or ϕ(b) Ω ∈ Θ.

  s), (t, s), (t, t)} , 1 if (x ′ 2p , x 2p+1 ) = (s, t) ,

	
	 
	ε 2,a =
	 

-1 if (x 1 , x ′ 2p , x 2p+1 ) ∈ {(s, s, s), (s, t, s), (s, t, t)} , 0 if (x 1 , x ′ 2p , x 2p+1 ) ∈ {(s, s, t), (t, s, s), (t, t, s), (t, t, t)} , 1 if (x 1 , x ′ 2p , x 2p+1 ) = (t, s, t) .

  -1 if (y 1 , y ′ 2p , y 2p+1 , y ′ 2p+1 ) ∈ {(s, s, s, s), (s, t, s, s), (s, t, t, s)} , 0 if (y 1 , y ′ 2p , y 2p+1 , y ′ 2p+1 ) ∈ {(s, s, s, t), (s, s, t, s), (s, t, s, t), (s, t, t, t), (t, s, s, s), (t, t, s, s), (t, t, t, s)} , 1 if (y 1 , y ′ 2p , y 2p+1 , y ′ 2p+1 ) ∈ {(s, s, t, t), (t, s, s, t), (t, s, t, s), (t, t, s, t), (t, t, t, t)} , 2 if (y 1 , y ′ 2p , y 2p+1 , y ′ 2p+1 ) = (t, s, t, t) ,

	
	 
	  
	  
	ε 2,b =
	  
	   
	
	  	-1 if (y ′ 2p , y 2p+1 , y ′ 2p+1 ) ∈ {(s, s, s), (t, s, s), (t, t, s)} ,
	ε 3,b =	0	if (y ′ 2p , y 2p+1 , y ′ 2p+1 ) ∈ {(s, s, t), (s, t, s), (t, s, t), (t, t, t)} ,
	 	1	if (y ′ 2p , y 2p+1 , y ′ 2p+1 ) = (s, t, t) ,
	
	 
	ε 4,b =
	 

′ 2p , y 2p+1 ) ∈ {(s, s), (t, s), (t, t)} , 1 if (y ′ 2p , y 2p+1 ) = (s, t) , -1 if (y 1 , y ′ 2p , y 2p+1 ) ∈ {(s,

s, s), (s, t, s), (s, t, t)} , 0 if (y 1 , y ′ 2p , y 2p+1 ) ∈ {(s, s, t), (t, s, s), (t, t, s), (t, t, t)} , 1 if (y 1 , y ′ 2p , y 2p+1 ) = (t, s, t) .

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• a p+1 b p+1 = c;

• We set x i = τ (a i ), x ′ i = σ(a i ), y i = σ(b i ), y ′ i = τ (b i ) for all i ∈ {1, . . . , p + 1}. Then x ′ i = x i+1 for all i ∈ {1, . . . , p -1}. We denote by ϕ : M → M the isomorphism that sends s to t and t to s. Since a i b i = ∆, we have y i = ϕ(x i ) and y ′ i = ϕ(x ′ i ) for all i ∈ {1, . . . , p}. In particular, y ′ i = ϕ(x ′ i ) = ϕ(x i+1 ) = y i+1 for all i ∈ {1, . . . , p -1}. Let u = |{i ∈ {1, . . . , p} | x ′ i = s}|. By Lemma 5.7, dpt(a) = dpt(a p+1 ) + p i=1 dpt(a i )u + ε a , where ε a is as follows. If p ≥ 1 and a p+1 = 1, then: ε a = 0 if (x ′ p , x p+1 ) ∈ {(s, s), (t, s), (t, t)} and ε a = 1 if (x ′ p , x p+1 ) = (s, t). If p ≥ 1 and a p+1 = 1, then: Finally, by Lemma 5.7 (2), we have p i=1 (dpt(a i ) + dpt(b i )) = pk. On the other hand, since y ′ i = ϕ(x ′ i ) for all i ∈ {1, . . . , p}, we have u + v = p. Set ε = ε a + ε bε c . By the above we have dpt(c) = dpt(a) + dpt(b)p(k -1)ε and ε is as follows. If p ≥ 1, a p+1 = 1 and b p+1 = 1, then: ε = 0 if (x ′ p , x p+1 , y p+1 ) ∈ {(s, s, t), (t, t, s)} and ε = 1 otherwise. If p ≥ 1, a p+1 = 1 and b p+1 = 1, then: ε = 0 if (x ′ p , x p+1 ) = (s, s) and ε = 1 otherwise. If p ≥ 1, a p+1 = 1 and b p+1 = 1, then: ε = 0 if (x ′ p , y p+1 ) = (t, s) and ε = 1 otherwise. If p ≥ 1, a p+1 = 1 and b p+1 = 1, then ε = 1. If p = 0, a = 1 and b = 1, then: ε = 0 if (x p+1 , y p+1 ) ∈ {(s, t), (t, s), (t, t)} and ε = 1 otherwise. If p = 0 and either a = 1 or

By Theorem 3.2 this implies the following.

Corollary 6. [START_REF] Burckel | The wellordering on positive braids[END_REF] The pair (H, G 1 ) is a Dehornoy structure on G.

We denote by P 1 the set of (H, G 1 )-positive elements of G and we set

Then by Proposition 3.1 we have the following. Corollary 6. [START_REF] Crisp | Injective maps between Artin groups, Geometric group theory down under[END_REF] The set P ǫ is the positive cone for a left-order on G.

Let r 1 , . . . , r 2k be the standard generators of the braid group B 2k+1 on m = 2k + 1 strands. Again, by Crisp [START_REF] Crisp | Injective maps between Artin groups, Geometric group theory down under[END_REF], we have an embedding ι : G → B 2k+1 which sends s to k-1 i=0 r 2i+1 and t to k i=1 r 2i . The proof of the following is substantially the same as the proof of Proposition 5.4, hence it is left to the reader.

We start now the proof of Theorem 6.1. We say that an element a ∈ M is Ω-unmovable if Ω ≤ L a or, equivalently, if Ω ≤ R a. The following is an observation. Lemma 6.5 [START_REF] Brieskorn | Artin[END_REF] Let a be an Ω-unmovable element of M . Then a is uniquely written in the form a = t up s vp • • • t u 1 s v 1 t u 0 , where u 0 , u p ≥ 0, u 1 , . . . , u p-1 ≥ 1 and v 1 , . . . , v p ≥ 1. In this case we have dpt(a) = p.

(2) Let a be an unmovable element of M . Then a is uniquely written in the form a = a ′ Ω ε where a ′ is Ω-unmovable and ε ∈ {0, 1}.

The first part of Theorem 6.1 is a direct consequence of this lemma.

Proposition 6. [START_REF] Dehornoy | Alternating normal forms for braids and locally Garside monoids[END_REF] The pair (H, G 1 ) satisfies Condition A with constant ζ = 2k -1.

Proof Let p ≥ 1 be an integer. We have θ = (st) k (ts) k , hence θ p = ((st) k (ts) k ) p . By Lemma 6.5 (1) it follows that dpt(

The second part of Theorem 6.1 will be much more difficult to prove. Let a ∈ M \ {1} be an Ω-unmovable element that we write as in Lemma 6.5 [START_REF] Brieskorn | Artin[END_REF]. Then we set σ(a) = t if u p = 0 and σ(a) = s if u p = 0. Similarly, we set τ (a) = t if u 0 = 1 and τ (a) = s if u 0 = 0. In other words, σ(a) is the first letter of a and τ (a) is its last one. On the other hand, we denote by ϕ : G → G the automorphism which sends s to t and t to s. Note that ϕ is the conjugation by Ω, that is, ϕ(α) = ΩαΩ -1 for all α ∈ G. The following is again a direct consequence of Lemma 6.5.

If p ≥ 1 and b 2p+1 = 1, then:

Again, by applying Lemma 6.7 we prove successively the following equalities.

If a 2p+1 = 1 and b 2p+1 = 1, then:

From Lemma 6.8 we also get 2p i=1 (dpt

Suppose that a ∈ Θ. Then a is written a = θ q with q ≥ 1. On the other hand we write

Suppose that aΩ ∈ Θ. Then aΩ is written aΩ = θ q t with q ≥ 1, hence a = θ q-1 (st) k . On the other hand we write b = s r b ′ , where b

On the other hand we write a = a ′ s r where either a ′ = 1 or Suppose that a ∈ Θ. Then a is written a = θ q with q ≥ 1. On the other hand we write b = t r b ′ where either b ′ = 1 or σ(b ′ ) = s. If r = 0 and b ′ = 1, then p = 0,

Then bΩ is written bΩ = θ q t with q ≥ 1, hence b = θ q-1 (st) k . On the other hand we write a = a ′ t r where a ′ = 1 (since a ∈ Θ) and τ (a ′ ) = s. If Suppose that aΩ ∈ Θ. Then aΩ is written aΩ = θ q t with q ≥ 1, hence a = θ q-1 (st) k . On the other hand we write b = s r b ′ , where either b

Then ϕ(b)Ω is written ϕ(b)Ω = θ q t with q ≥ 1, hence b = (ts) k θ q-1 . On the other hand we write a = a ′ s r where either a ′ = 1 or τ (a ′ ) = t. If r = 0 and a ′ = 1, then p = 0, a = 1, b = (ts) k θ q-1 = 1 and y 2p+1 = t, hence ε = 1. If r = 0 and a ′ = 1, then p = 0, a = a ′ = 1, b = (ts) k θ q-1 = 1 and (x 2p+1 , y 2p+1 ) = (t, t), hence ε = 1. If r > 0 and a ′ = 1, then r = 2p ≤ 2(q -1),

Lemma 6.10 Let a, b be two Ω-unmovable elements of M . We assume that the ∆-form of ab is in the form ab = (cΩ)∆ p where c is an Ω-unmovable element of M and p ≥ 0.

(1) Suppose that (a, b) ∈ ( Θ × Θ). There exists ε ∈ {0, 1} such that dpt(cΩ) = dpt(a)

Proof We write a and b in the form

i for all i ∈ {1, . . . , 2p}. For i ∈ {1, . . . , 2p + 1} we have y i = ϕ(x i ) and y ′ i = x ′ i if i is odd and y i = x i and

By using Lemma 6.7 we obtain successively the following equalities.

, where ε 1,a and ε 2,a are as follows. If a 2p+2 = 1, then:

Similarly, by using Lemma 6.7 we obtain successively the following equalities.

where ε 1,b , ε 2,b , ε 3,b and ε 4,b are as follows. If b 2p+2 = 1, then:

, (t, s, s, s), (t, t, s, s), (t, t, t, s)} ,

Again, by using Lemma 6.7 we obtain the following equalities.

where ε 1,c and ε 2,c are as follows. If a 2p+2 = 1 and b 2p+2 = 1, then:

If a 2p+2 = 1 and b 2p+2 = 1, then:

If a 2p+2 = 1 and b 2p+2 = 1, then:

Finally, from Lemma 6.7 and Lemma 6.8 follows that Suppose that aΩ ∈ Θ. Then aΩ is written aΩ = θ q t with q ≥ 1, hence a = θ q-1 (st) k . On the other hand we write b = s r b ′ where b

Suppose that ϕ(b) ∈ Θ. Then ϕ(b) is written ϕ(b) = θ q with q ≥ 1, hence b = ((ts) k (st) k ) q . On the other hand we write a = a ′ s r where either a ′ = 1 or τ (a ′ ) = t.

We necessarily have r = 2p + 1 < 2q, a 2p+2 = a ′ and b 2p+2 = ((ts) k (st) k ) q-p-1 (ts) k , hence x ′ 2p+1 = s and x 2p+2 = t if a ′ = 1, and therefore ε = 1. Suppose that a ∈ Θ. Then a is written a = θ q with q ≥ 1. On the other hand we write b = t r b ′ where either b ′ = 1 or σ(b ′ ) = s. We necessarily have r = 2p+1 < 2q, hence

Suppose that bΩ ∈ Θ. Then bΩ is written bΩ = θ q t with q ≥ 1, hence b = θ q-1 (st) k .

On the other hand we write a = a ′ t r where a ′ = 1 (since a ∈ Θ) and τ (a Suppose that aΩ ∈ Θ. Then aΩ is written aΩ = θ q t with q ≥ 1, hence a = θ q-1 (st) k . On the other hand we write b = s r b ′ where either b ′ = 1 or σ(b ′ ) = t. If r ≥ 2q -1, then a 2p+2 = 1 and x ′ 2p+1 = s, hence ε = 1. If r < 2q -1, then r = 2p + 1, a 2p+2 = θ q-p-1 and b 2p+2 = b ′ , hence x ′ 2p+1 = s, x 2p+2 = s and either b 2p+2 = 1 or y 2p+2 = t, and therefore ε = 1.

Suppose that ϕ(b)Ω ∈ Θ. Then ϕ(b)Ω is written ϕ(b)Ω = θ q t with q ≥ 1, hence b = ((ts) k (st) k ) q-1 (ts) k . On other hand we write a = a ′ s r where either a ′ = 1 or τ (a ′ ) = t. If r ≥ 2q -1, then b 2p+2 = 1 and x ′ 2p+1 = s, hence ε = 1. If r < 2q -1, then r = 2p + 1, a 2p+2 = a ′ and b 2p+2 = ((ts) k (st) k ) q-p-1 , hence x ′ 2p+1 = s and y 2p+2 = t, and therefore ε = 1. Now, the second part of Theorem 6.1 is a direct consequence of the previous two lemmas.