Mireille Soergel
email: mireille.soergel@u-bourgogne.fr

Systolic complexes and group presentations

come

Introduction

In order to better understand a group, one approach is to understand whether it acts properly discontinuously and cocompactly by isometries on a non-positively curved space. For Riemannian manifolds, one could consider the sectional curvature of the manifold. More generally for geodesic metric spaces, we see if they satisfy the CAT(0) inequality. For simplicial complexes Januszkiewicz and Swiatkowski introduced in [START_REF] Januszkiewicz | Simplicial nonpositive curvature[END_REF] the notion of systolic complexes as a combinatorial form of non positive curvature. Though their 1-skeleta had been studied earlier by Chepoi under the name bridged graphs [START_REF] Chepoi | Graphs of some CAT(0) complexes[END_REF]. A flag simplicial complex is systolic if it is simply connected and all of its vertex links are 6-large, i.e. all cycles of length 4 or 5 have diagonals. The main consequences for a group G acting properly discontinuously and cocompactly on a systolic complex are the following:

1. G is biautomatic. This was already proven by Januszkiewicz and Swiatkowski in their original paper on systolicity [START_REF] Januszkiewicz | Simplicial nonpositive curvature[END_REF]. This especially implies solvability of the Word Problem and of the Conjugacy Problem and a quadratic Dehn function.

3. Virtually solvable subgroups of G are either virtually cyclic or virtually Z 2 . This is mentioned in [START_REF] Huang | Large-type Artin groups are systolic[END_REF]. Bestvina showed that virtually solvable subgroups of spherical Artin groups are abelian [START_REF] Bestvina | Non-positively curved aspects of Artin groups of finite type[END_REF]. Gersten-Short showed a result for polycyclic subgroups of biautomatic groups [START_REF] Gersten | Rational subgroups of biautomatic groups[END_REF].

4. The centralizer of an infinite-order element of G is commensurable with F n × Z or Z [START_REF] Crisp | Automorphisms and abstract commensurators of 2dimensional Artin groups[END_REF], [START_REF] Elsner | Isometries of systolic spaces[END_REF].

As a group naturally acts on its Cayley graph, this leads to the following questions: How do we construct a flag simplicial complex from the Cayley graph? Can we give conditions on the presentation which ensure that this space is systolic? The goal of this paper is to give a partial answer to these questions. Systolicity is already known for large type Artin groups [START_REF] Huang | Large-type Artin groups are systolic[END_REF] and right angled Artin groups with bipartite defining graph [START_REF] Elsner | Square complexes and simplicial nonpositive curvature[END_REF]. Triangular Coxeter groups (except (2,4,4),(2,4,5),(2,5,5)) are also systolic [START_REF] Przytycki | Systolizing buildings[END_REF]. In this paper we will focus our applications on Garside groups and Artin groups. We will first give some background on systolic complexes and set some conditions on group presentation that lead to a simplicial Cayley complex with free group action. We introduce the notion of a restricted triangular presentation. We say that a presentation S | R of some groupe G is a restricted triangular presentation if S ∩ S -1 = ∅, R = {a • b • c -1 | abc -1 = e in G} and for a, b, c ∈ S abc ∈ S ⇒ ab, bc ∈ S. In this case the Cayley complex is simplicial. We then establish when the flag complex of such a Cayley complex is sytolic. To do so we study the cycles of length four and five. This leads to Theorem 1. (Theorem 3.5) Consider a group G with generating set S, where G has a finite restricted triangular presentation with respect to S. Then the complex F lag(G, S) is a simply connected simplicial complex. It is systolic if and only if the generating set S satisfies the following conditions:

1) If ∃u, w, a, b, c, d ∈ S, u = w, a = d, with ua = wb ∈ S and ud = wc ∈ S then ∃k ∈ S such that w = uk or ua = udk.

2) If ∃v, x, a, b, c, d ∈ S, v = x, a = b, with bv = cx ∈ S and av = dx ∈ S then ∃k ∈ S such that v = kx or av = kbv. Moreover, this implies that G is a systolic group.

We say that a presentation satisfying the conditions of Theorem 1 is a systolic presentation. We present some examples which show that these conditions are all necessary.

There are two main applications of Theorem 1. The first concerns Garside groups. They were introduced by Dehornoy and Paris in [START_REF] Dehornoy | Gaussian groups and Garside groups, two generalisations of Artin groups[END_REF] as a generalization of spherical Artin groups. The Garside structure on a group naturally gives a presentation leading to a simplicial Cayley graph, we call this the Garside presentation of a Garside group. We can classify the Garside groups whose Garside presentation satisfy Theorem 3.5. Consider the following definitions. Let x 1 , . . . , x n be n letters and let m be a positive integer. We define prod(x 1 , . . . , x p ; m) = x 1 x 2 . . . x p x 1 x 2 . . .

G n,m = x 1 , . . . , x n | prod(x 1 , . . . , x n ; m) = prod(x 2 , . . . , x n , x 1 ; m) = . . . = prod(x n , x 1 , . . . , x n-1 ; m) .
We can then make the following statement about systolic Garside groups:

Theorem 2. (Theorem 4.6) Let G be a Garside group of finite type. Then G has a systolic Garside presentation if and only if G = ( * p i=1 G n i ,m i )/(∆ n i ,m i = ∆ n j ,m j ∀i, j) for some positive integers p, n 1 , . . . , n p and m 1 , . . . , m p .

The second application concerns Artin groups. To state the next result we require a few definitions. An orientation on a simple graph Γ is an assignement o(e) for each edge e ∈ E(Γ) where o(e) is a set of one or two endpoints of e. An edge with both endpoints assigned is bioriented. The startpoint i(e) is an assignement of one or two startpoints of e, which is consistent with the choice of o(e). If o(e) consists of one point, i(e) consists of one point such that e = (i(e), o(e)), if o(e) consists of two points then so does i(e). We say that a cycle γ is directed if for each v ∈ γ there is exactly one edge e ∈ γ with v ∈ o(e). A cycle is undirected if it is not directed. We say that a 4

-cycle γ = (a 1 , a 2 , a 3 , a 4 ) is misdirected if a 2 ∈ o(a 1 , a 2 ), a 2 ∈ o(a 2 , a 3 ), a 4 ∈ o(a 3 , a 4 ) and a 4 ∈ o(a 4 , a 1 ).
A Γ = s v , v ∈ V | [s v s w s v . . . ] me = [s w s v s w . . . ] me
for all edges e = (v, w) with label m e .

We can now present the following result: Theorem 3. (Theorem 4.14) Let Γ be a simple graph, with edges labeled by numbers ≥ 2 and with an orientation o such that an edge is bioriented if and only if it has label 2. Assume that every 3-cycle is directed and no 4-cycle is misdirected. Let A be the Artin group associated to Γ. Then the dual presentation of A Γ induced by this orientation is systolic.

Preliminaries

We start with some background on simplicial and systolic complexes. Let X be a simplicial complex. Assume it is finite dimensional and locally finite. We denote its k-skeleton by X (k) . Then X (0) is the set of vertices of X. The subcomplex spanned by A ⊂ X (0) is the largest subcomplex of X which has A as its set of vertices. The complex X is flag if every set of pairwise adjacent vertices spans a simplex. A flag complex is uniquely determined by its 1-skeleton. For a simplex σ ∈ X we can define its link in X,

Lk(σ, X) = {τ ∈ X | τ ∩ σ = ∅ and τ ∪ σ ∈ X}.
A cycle in X is the image of a simplicial map f : S 1 → X from a triangulation of the 1-sphere to X. If f is injective, the cycle is embedded. Let γ be an embedded cycle in X. The length of γ, |γ|, is the number of edges of γ. We say that γ is a |γ|-cycle. A diagonal of γ is an edge that connects two nonconsecutive vertices of γ. An embedded cycle is diagonal free if there are no edges between nonconsecutive vertices. We say that two vertices v and w are adjacent if there exists an edge between v and w, we then write v ∼ w. A simplicial complex is 6-large if every embedded cycle γ with 4 ≤ |γ| < 6 has a diagonal. A simplicial complex is systolic if it is connected, simply connected and if Lk(v, X) is flag and 6-large for all vertices v ∈ X. A group is systolic if it acts properly discontinuously and cocompactly on a systolic complex. To know when a Cayley complex is systolic we first need to determine when it is simplicial.

Let G be a group and S ⊂ G a finite set of generators. Suppose additionally that S ∩ S -1 = ∅, this especially implies e / ∈ S and s 2 = e for all s ∈ S. Let Γ(G, S) be the Cayley graph of G relative to S. Its vertices and edges are

V (Γ(G, S)) = {v[g] | g ∈ G} and E(Γ(G, S)) = {e[g, s] | g ∈ G, s ∈ S} where the edge e[g, s] goes from v[g] to v[gs].
We also write e[g, s] = (v[g], v[gs]). As S ∩ S -1 = ∅, the graph Γ(G, S) is simplicial. So we can define F lag(G, S) as the flag complex of Γ(G, S). As F lag(G, S) is the flag complex of Γ(G, S), the group G naturally acts properly discontinuously and cocompactly by isometries on F lag(G, S).

Proposition 2.1. Let G be a group and S ⊂ G a finite generating set. Suppose additionally that S ∩ S -1 = ∅. Then F lag(G, S) is a simply connected simplicial complex and π 1 (F lag(G, S)/G) = G if and only if G admits the presentation

G = S | R where R = {a • b • c | a, b, c ∈ S with abc = e in G} ∪ {a • b • c -1 | a, b, c ∈ S with abc -1 = e in G}.
Proof. To see when F lag(G, S) is simply connected it is enough to take a look at its 2-skeleton. There is a 2-simplex in F lag(G, S) for every set of 3 pairwise adjacent vertices. As in Γ(G, S) edges are labeled by elements in S and vertices correspond to elements of G, so are edges and vertices in F lag(G, S). Hence we can interpret the existence of a 2-simplex in terms of relations on the generators. At each vertex v[g] in F lag(G, S) and for every triple a, b, c ∈ S with a for which one of these equalities holds in G .

• b • c = e or a • b • c -1 = e in G there is a 2-simplex with vertices v[g], v[ga], v[gab]
Finally F lag(G, S) is simply connected if and only if it is the universal cover of the quotient, so if and only if π 1 (F lag(G, S)/G) = G.

We call a presentation satisfying the conditions of Proposition 2.1 a triangular presentation.

Proposition 2.2. Assume a group G has a triangular presentation

S | R . As- sume additionally R = {a • b • c -1 | a, b, c ∈ S with abc -1 = e in G}.
Then the action of G on F lag(G, S) is free.

Proof. Let g ∈ G and x ∈ F lag(G, S) such that g • x = x. Let V be the set of vertices of the smallest simplex containing x. As the action of G is simplicial, g • V = V . The restriction of possible relations in R imposes an orientation on triangles in the graph, which in turn implies that in the subcomplex V there exists a unique vertex v 0 ∈ V with only incoming edges i.e. ∃v 0 ∈ V such that ∀w ∈ V \ {v 0 }, ∃k ∈ S such that v 0 = wk. Since S ∩ S -1 = ∅, the action of G on Γ(G, S) preserves the orientation of the edges and is free on the vertices. So g

• x = x implies g • v 0 = v 0 hence g = e.

Systolic Cayley Complexes

Consider a group G with a finite triangular presentation G = S | R . Then we know that F lag(G, S) is a simply connected simplicial complex. We now want to know when F lag(G, S) is systolic. We already know that F lag(G, S) is simply connected. As it is flag, all the links of vertices are flag. So we need to check whether Lk(v, F lag(G, S)) is 6-large for all vertices of F lag(G, S). As the action of G on F lag(G, S) is transitive and by isometries on the vertices, we only need to check if Lk(e, F lag(G, S)) is 6-large. For simpler notation we set L = Lk(e, F lag(G, S)).

Then the vertices of

L are V (L) = {v[s] | s ∈ S ∪S -1 }.
As we differentiate between elements in S and elements in S -1 , we will call vertices v[s] with s ∈ S positive and vertices v[s] with s ∈ S -1 negative. Edges between these vertices are labeled by elements in S and

E(L) = {e[g, a] | g ∈ S ∪S -1 and a ∈ S \{g -1 } and ga ∈ S ∪S -1 }, where e[g, a] is the edge going from v[g] to v[ga].
We write v ∼ w for two adjacent vertices v and w and e = (v, w) for the edge e between v and w. If there is an edge from v to w we might also use the notation v → w. To simplify the notation, we will also denote the vertex v[s] with s and say the vertex s ∈ S is positive, s ∈ S -1 is negative.

We put some additional conditions on S and R:

1. If for some a, b, c ∈ S, abc ∈ S then ab ∈ S and bc ∈ S.

2. R = {a • b • c -1 | abc -1 = e in G}, so we do not have relations of the form a • b • c in R.
We call a triangular presentation satisfying these additional conditions a restricted triangular presentation. These conditions are mostly technical. They limit the possible diagonal free cycles in L and by Proposition 2.2 ensure that the action of G on F lag(G, S) is free. We don't know how to decide whether L is 6-large without them. So what can we say about diagonal free cycles in L of length 4 or 5 under these conditions?

Remark 3.1. The additional condition on R implies that there are no edges from vertices in S to vertices in S -1 , so from positive to negative vertices. The additional conditions on S imply that every cycle γ in L which contains one of the following configurations of adjacent vertices has a

diagonal. So if a, b, c ∈ S with a) a, b, c ∈ V (γ) and a → b → c then a → c. b) a -1 , b -1 , c -1 ∈ V (γ) and a -1 → b -1 → c -1 then a -1 → c -1 . c) a -1 , b, c ∈ V (γ) and a -1 → b ← c then a -1 → c. d) a -1 , b -1 , c ∈ V (γ) and a -1 ← b -1 → c then a -1 → c.
So all of these configurations of vertices cannot occur in diagonal free cycles. This leads us to the following statement about potential cycles of length 4 or 5.

Lemma 3.2. Every cycle of length 5 in L contains a diagonal.

Proof. Let γ be a cycle of length 5 in L.

As 5 is odd, if γ has 5 positive or 5 negative vertices, it has a diagonal (situation a) or b) always occurs).

Assume γ contains one negative and four positive vertices. As there are only edges from negative to positive vertices, the direction of two of the edges in γ is already determined. Every possible direction of the three other edges leads to one of the situations above (avoiding situation c) necessarily leads to situation a)). The same argument holds if γ has one positive and four negative vertices.

Assume γ has two negative and 3 positive vertices. As each negative vertex is adjacent to at least one positive vertex, the direction of at least two edges of γ is already determined. Every possible direction of the three other edges leads to one of the situations above (if the two negative vertices are adjacent we have situation d) otherwise situation c)). The same argument holds if γ has two positive and three negative vertices.

Lemma 3.3. Let u, v, w, x, a, b, c, d ∈ S. The only cycles of length 4 in L that do not contain one of the situations mentioned in Remark 3.1 are:

1) 2) 3) 4) 5) u d x w c a v b u -1 d x -1 w -1 c a v -1 b u -1 d x c w a v b u -1 d x -1 c w a v -1 b u -1 d x w -1 c a v b
Proof. Let γ be a cycle of length 4. Assume γ has 4 positive vertices. In order not to be in situation Remark 3.1 a) each vertex has either two incoming or two outgoing vertices. This corresponds to the first cycle. The same argument holds if γ has 4 negative vertices. Then we have the second cycle.

Assume γ has 1 negative and 3 positive vertices. Then the negative vertex is adjacent to two positive vertices and hence the direction of two edges is already determined. As we do not allow the configuration Remark 3.1 c) the only possible cycle is the third cycle. The same argument holds for 1 positive and 3 negative vertices, which gives the fourth cycle.

Assume γ has 2 positive and 2 negative vertices. Then if the two negative vertices are adjacent we are necessarily in situation Remark 3.1 d). So the negative vertices are not adjacent. Hence they are both adjacent to the two positive vertices and the direction of these edges is determined. This gives the fifth cycle.

So to see if L is 6-large we need to concentrate on the cycles of length 4 presented in Lemma 3.3. When do they exist? Under which conditions do they have a diagonal? The next lemma aims to answer those questions. Note that u, v, w, x ∈ S correspond to vertices in 4-cycles in L and a, b, c, d ∈ S to edges. Also note that these conditions are all necessary as one can see in Lemma 3.6.

Proof. We know by Lemma 3.2 that there are no diagonal free cycles of length 5 in L. By Lemma 3.3, we know that there are only five problematic cycles of length 4. We show here under which conditions on S such cycles exists and which conditions are necessary for the existence of a diagonal. If those five 4-cycles have a diagonal, all cycles of length 4 have a diagonal. So all cycles of length < 6 have a diagonal, so L is 6-large. The existence of the 4-cycle relies on two elements: we need 4 distinct vertices u, v, w, x and we need the appropriate edges a, b, c, d between these vertices. i) In the first cycle of length 4: the existence of the cycle is equivalent to the following statement about elements of S:

∃ a, b, c, d, u, v, w, x ∈ S, u, v, w, x pairwise distinct such that v = ua = wb and x = ud = wc where u, v, w, x ∈ S are the labels on the vertices and a, b, c, d are the labels on the edges. There is a diagonal if v ∼ x or u ∼ w which is equivalent to

∃ k ∈ S : w = uk, u = wk, v = xk or x = vk
where k ∈ S is the label on the diagonal. Using the symmetries of the 4-cycle, this corresponds to condition 1).

ii) In the second cycle of length 4: the existence of the cycle is equivalent to the following statement about elements of S:

∃ a, b, c, d, u, v, w, x ∈ S, u -1 , v -1 , w -1 , x -1 pairwise distinct such that v -1 = u -1 a = w -1 b and x -1 = w -1 c = u -1 d
where a, b, c, d ∈ S are the labels on the edges. There is a diagonal if

v -1 ∼ x -1 or u -1 ∼ w -1 which is equivalent to ∃ k ∈ S : w -1 = u -1 k, u -1 = w -1 k, v -1 = x -1 k or x -1 = v -1 k
where k ∈ S is the label on the diagonal. Using the symmetries of the 4-cycle this corresponds to condition 2).

iii) In the third cycle of length 4: the existence of the cycle is equivalent to the following statement about elements of S:

∃ a, b, c, d, u, v, w, x ∈ S, u -1 , v, w, x pairwise distinct such that v = u -1 a, x = u -1 d, w = vb and w = xc
where a, b, c, d ∈ S are the labels on the edges. There is a diagonal if v ∼ x or u -1 ∼ w which is equivalent to

∃k ∈ S : uw = k, v = xk or x = vk
where k ∈ S is the label on the diagonal. As u -1 is a negative vertex and w is a positive one, there is only one possible direction for the diagonal from u -1 to w. Using the symmetry of the 4-cycle this corresponds to condition 3). iv) In the fourth cycle of length 4: the existence of the cycle is equivalent to the following statement about elements of S:

∃ a, b, c, d, u, v, w, x ∈ S, u -1 , v -1 , w, x -1 pairwise distinct such that v -1 = u -1 a, w = v -1 b = x -1 c and x -1 = u -1 d
where a, b, c, d ∈ S are the labels on the edges. There is a diagonal if

v -1 ∼ x -1 or u -1 ∼ w which is equivalent to ∃k ∈ S : uw = k, v -1 = x -1 k or x -1 = v -1 k
where k ∈ S is the label on the diagonal. As u -1 is a negative vertex and w is a positive one, there is only one possible direction for the diagonal from u -1 to w. Using the symmetry of the 4-cycle this corresponds to condition 4).

v) In the fifth cycle of length 4: the existence of the cycle is equivalent to the following statement about elements of S:

∃ a, b, c, d, u, v, w, x ∈ S, u -1 , v, w -1 , x pairwise distinct such that v = u -1 a = w -1 b and x = u -1 d = w -1 c
where a, b, c, d ∈ S are the labels on the edges. There is a diagonal if v ∼ x or u -1 ∼ w -1 which is equivalent to

∃ k ∈ S : w -1 = u -1 k, u -1 = w -1 k, v = xk or x = vk
where k ∈ S is the label on the diagonal. Using the symmetries of the 4-cycle this corresponds to condition 5).

We can now get back to the original question: when is F lag(G, S) systolic?

Theorem 3.5. Consider a group G with generating set S, where G has a finite restricted triangular presentation with respect to S. Then the complex F lag(G, S) is a simply connected simplicial complex. It is systolic if and only if the generating set S satisfies the following conditions: Moreover, this implies that G is a systolic group.

Proof. It follows from Proposition 2.1 and the definition of a restricted triangular presentation that Γ(G, S) is a connected simplicial graph and F lag(G, S) is a welldefined simply connected flag complex. The complex F lag(G, S) is systolic if and only if the link of every vertex is flag and 6-large. As F lag(G, S) is a flag complex, the link of a vertex is flag. The action of G on the vertices of Γ(G, S) = F lag(G, S) (1) is transitive and by isometries. So F lag(G, S) is systolic if and only if the link L =Lk(e, F lag(G, S)) is 6-large. This is equivalent to the conditions given by Lemma 3.4. Since G acts properly discontinuously and cocompactly on F lag(G, S), the group G is systolic if F lag(G, S) is systolic.

We call a presentation satisfying the conditions of Theorem 3.5, a systolic presentation. We say a group is Cayley systolic if it admits a systolic presentation. By Proposition 2.2 a Cayley systolic group G with systolic presentation S | R acts freely on F lag(G, S). One can also note that free products of Cayley systolic groups are also Cayley systolic. More generally we do not know under which conditions amalgamated products of Cayley systolic groups are systolic or Cayley systolic.

Lemma 3.6. Consider the set S = {a, b, c, d, u, v, w, x} and the sets

R 1 = {uav -1 , wbv -1 , udx -1 , wcx -1 }, R 2 = {bvw -1 , cxw -1 , avu -1 , dxu -1 }, R 3 = {vbw -1 , xcw -1 , uxd -1 , uva -1 }, R 4 = {dxu -1 , avu -1 , vwb -1 , xwc -1 }, R 5 = {vua -1 , vwb -1 , xwc -1 , xud -1 }.
For all i ∈ {1, 2, 3, 4, 5}, the presentation S | R i is a restricted triangular presentation. Additionally the presentation S | R i satisfies all the conditions of Theorem 3.5 except condition i.

Proof. We show this for S | R 3 . The other cases can be checked in the same way. First note that a = uv, b = v -1 w, c = x -1 w and d = ux. So the group

C 3 = S | R 3 is in fact F (u, v, w, x
) the free group on the generators u, v, w and x. So the word problem in C 3 is solvable. We do the following calculations using SageMath. We check for all triples (α, β, γ) ∈ S 3 that αβγ = 1, αβγ / ∈ S and αβγ -1 = e ⇔ α • β • γ -1 ∈ R 3 . Note that αβγ / ∈ S implies S ∩ S -1 = ∅. So the presentation S | R 3 is a restricted triangular presentation. Now we need to check the different conditions of Theorem 3.5. Using SageMath, we see that there are no elements s u , s v , s w , s x , s a , s b , s c , s d ∈ S satisfying the hypothesis of one of the conditions 1), 2), 4) and 5) but the tuples (s u , s v , s x , s b , s c ) = (u, v, x, b, c) and

(s u , s v , s x , s b , s c ) = (u, x, v, c, b) satisfy s v s b = s x s c ∈ S, s u s x ∈ S and s u s v ∈ S.
We check that condition 3) fails in at least one of those cases so we check that for all s ∈ S we have s = s u s v s b and s v = s x s for at least one of those tuples.

Example 3.7. The conditions in Theorem 3.5 are all necessary as shown in Lemma 3.6. One can note that the given presentations are not systolic but the underlying groups are, since free groups are known to be systolic and even Cayley systolic with respect to the standard generating system. Also note that the following presentation

F 2 × F 2 = a, b, c, d, ∆ 1 , ∆ 2 , ∆ 3 , ∆ 4 | ∆ 1 = ab = ba, ∆ 2 = bc = cb, ∆ 3 = cd = dc, ∆ 4 = da = ad
satisfies conditions 1)-4) but not condition 5). Again F 2 × F 2 is known to be systolic, using a construction by Elsner and Przytycki [START_REF] Elsner | Square complexes and simplicial nonpositive curvature[END_REF]. We do not know whether it is Cayley systolic.

Remark 3.8. We can also interpret the conditions in Lemma 3.4 as conditions on a given order on S. We first introduce left and right orders on S ∪ e by defining for a, b ∈ S ∪ {e}:

a) a ≤ L b if ∃c ∈ S ∪ {e} such that ac = b. b) a ≤ R b if ∃c ∈ S ∪ {e} such that ca = b.
These are indeed orders on S ∪ {e}: as e ∈ S ∪ {e} they are reflexive, as there are no inverses in S they are antisymmetric and by the additional condition on S they are transitive. An edge v → w in L is equivalent to

1. v ≤ L w if v, w ∈ S 2. w -1 ≤ R v -1 if v, w ∈ S -1 3. v -1 w ∈ S if w ∈ S and v ∈ S -1 .
Then the conditions on the existence of diagonals in given 4-cycles could be reformulated in the following way:

1) If ∃u, v, w, x ∈ S pairwise distinct with u ≤ L v and w ≤ L v and u ≤ L x and w ≤ L x then u ≤ L w or x ≤ L v.

2) If ∃u, v, w, x ∈ S pairwise distinct with v ≤ R u and x ≤ R u and v ≤ R w and x ≤ R w then u ≤ R w or v ≤ R x.

3) If ∃u, v, w, x ∈ S pairwise distinct with uv ∈ S and ux ∈ S and v ≤ L w and x ≤ L w then uw ∈ S or v ≤ L x. Here the conditions are only given in terms of elements of S associated to the vertices of the 4-cycles.

Applications

Garside groups

A group G is said to be a Garside group with Garside structure (G, P, ∆) if it admits a submonoid P with P ∩ P -1 = {e}, called the monoid of positive elements and a special element ∆ ∈ P called Garside element such that the following properties are satisfied:

1. The partial order ≤ L defined by a ≤ L b ⇔ a -1 b ∈ P is a lattice order, i.e. for every a, b ∈ G there exists a unique lcm a ∨ L b and a unique gcd a ∧ L b with respect to

≤ L i.e. ∀a, b ∈ G ∃!(a ∨ L b) such that a ≤ L (a ∨ L b), b ≤ L (a ∨ L b) and ∀c ∈ G a ≤ L c and b ≤ L c imply (a ∨ L b) ≤ L c. Similarly ∀a, b ∈ G ∃!(a ∧ L b) such that (a ∧ L b) ≤ L a, (a ∧ L b) ≤ L b and ∀c ∈ G c ≤ L a and c ≤ L b imply c ≤ L (a ∧ L b).
2. The set [e, ∆] = {a ∈ G | e ≤ L a ≤ L ∆}, called the set of simple elements, generates P .

3. Conjugation by ∆ preserves P i.e. ∆ -1 P ∆ = P .

4. For all x ∈ P \ {e} one has

x = sup{k ∈ N | ∃a 1 , . . . , a k ∈ P \ {e} such that x = a 1 • • • a k } < ∞.
A Garside structure (G, P, ∆) is said to be of finite type if the set of simple elements [e, ∆] is finite. A group G is said to be a Garside group of finite type if it admits a Garside structure of finite type. Elements x ∈ P \ {e} with x = 1 are called atoms. The set of atoms also generates P .

Remark 4.1. The monoid P also induces a partial order ≤ R which is invariant under right multiplication. We define a ≤ R b ⇔ ba -1 ∈ P . It follows from the properties of G that ≤ R is also a lattice order, that P is the set of elements a such that e ≤ R a and that the simple elements are the elements a such that e ≤ R a ≤ R ∆. We denote a ∨ R b the lcm and a ∧ R b the gcd with respect to ≤ R . So

{a ∈ G | e ≤ L a ≤ L ∆} = {a ∈ G | e ≤ R a ≤ R ∆}, we also say that ∆ is balanced.
Example 4.2. Spherical Artin groups are Garside groups, in particular braid groups are Garside. Torus knot groups x, y | x p = y q with p, q > 1 are Garside groups with Garside element ∆ = x p = y q . The fundamental group of the complement of n line through the origin in

C 2 x 1 , . . . , x n | x 1 x 2 . . . x n = x 2 . . . x n x 1 = • • • = x n x 1 . . . x n-1
is also a Garside group with Garside element ∆ = x 1 x 2 . . . x n .

Lemma 4.3. Let G be a Garside group with Garside element ∆ and set of simple elements S. Then G = S \ {e} | s • t = st ∀s, t ∈ S such that st ∈ S is a restricted triangular presentation of G.

Proof. This presentation is a direct consequence of Theorem 6.1 in [START_REF] Dehornoy | Gaussian groups and Garside groups, two generalisations of Artin groups[END_REF]. It is a restricted triangular presentation as all relations are of the form a • b = c for some a, b, c ∈ S and S is the set of simple elements of a Garside group.

We call this presentation a Garside presentation of G with Garside element ∆. We can now state one of our main results.

Theorem 4.4. Let G be a Garside group of finite type with Garside element ∆ and non trivial simple elements S. Then F lag(G, S) is systolic if and only if for all a, b ∈ S, a ∧ L b ∈ {e, a, b} and a ∧ R b ∈ {e, a, b}. In particular if F lag(G, S) is systolic then so is G.

Proof. By Lemma 4.3, the Garside presentation of G with generating set S is a restricted triangular presentation. By Proposition 2.1, F lag(G, S) is well-defined. By Theorem 3.5, F lag(G, S) is systolic if and only if the conditions 1)-5) are satisfied.

Assume that for all a, b ∈ S a ∧ L b, a ∧ R b ∈ {e, a, b}. Then 1) If ∃u, w, a, b, c, d ∈ S, u = w, a = b with ua = wb ∈ S and ud = wc ∈ S, then u ≤ L ua and u ≤ L ud so ua ∧ L ud = e, so either ua ∧ L ud = ua or ua ∧ L ud = ud, say ua ∧ L ud = ud so ua = udk for some k ∈ S.

2) If ∃v, x, a, b, c, d ∈ S, v = x, a = b with bv = cx ∈ S and av = dx ∈ S then v ≤ R bv and v ≤ R av then av ∧ R bv = e, so either av ∧ R bv = av or av ∧ R bv = bv, say av ∧ R bv = bv so av = kbv for some k ∈ S. So if for all a, b ∈ S, a ∧ R b, a ∧ L b ∈ {e, a, b}, the conditions 1)-5) of Theorem 3.5 are satisfied and so F lag(G, S) is systolic, which directly implies that G is sytolic.

We now show the other implication.

= r a k a = r b k b . Then c, k a , k b , r a , r b ∈ S and k a = k b , k a c = a ∈ S, k b c = b ∈ S and ∆ = r a k a = r b k b ∈ S. But r a k a c = ∆c / ∈ S and k ∈ S with k a = kk b or k b = kk a since c = a ∧ R b = a, b. So condition 4) of Theorem 3.5 fails. So F lag(G, S) is not systolic. So if there exist a, b ∈ S with a ∧ L b / ∈ {e, a, b} or a ∧ R b /
∈ {e, a, b}, the complex F lag(G, S) is not systolic.

Example 4.5. Let x 1 , . . . , x n be n letters and let m be a positive integer. We define

prod(x 1 , . . . , x p ; m) = x 1 x 2 . . . x p x 1 x 2 . . . m .
and prod(x 1 , . . . , x p ; 0) = e. Consider the group

G n,m = x 1 , . . . , x n | prod(x 1 , . . . , x n ; m) = prod(x 2 , . . . , x n , x 1 ; m) = . . . = prod(x n , x 1 , . . . , x n-1 ; m) .
By Proposition 5.2 in [START_REF] Dehornoy | Gaussian groups and Garside groups, two generalisations of Artin groups[END_REF], this is a Garside group with Garside element ∆ n,m = prod(x 1 , . . . , x n ; m). When considering all indices modulo n, we can write the set of simple elements as

S = {prod(x i , . . . , x i+n ; k) | 0 ≤ k ≤ m and 1 ≤ i ≤ n}.
In particular, for n = 1, we have G 1,m = x 1 with Garside element x m 1 and the simple elements are

S = {x i 1 | 0 ≤ i ≤ m}. Note that if m = n = 1 we have x 1 = ∆ 1,1 and S = {1, x 1 }. Also if n > 1 we can assume m > 1.
More generally for some positive intergers p, n 1 , . . . , n p , m 1 , . . . , m p , the product

G = ( * p i=1 G n i ,m i )/(∆ n i ,m i = ∆ n j ,m j ∀i, j) is a Garside group with Garside element ∆ = ∆ n 1 ,m 1 = • • • = ∆ np,mp .
We would like to remark that if p > 1 and n i = m i = 1 for some i, say i = p, we have

( * p i=1 G n i ,m i )/(∆ n i ,m i = ∆ n j ,m j ∀i, j) ∼ = ( * p-1 i=1 G n i ,m i )/(∆ n i ,m i = ∆ n j ,m j ∀i, j).
So we can assume that if p > 1, we have m i ≥ 2 for all i ∈ {1, . . . , p}. The next theorem shows that these are the only Garside groups with systolic Garside presentation.

Theorem 4.6. Let G be a Garside group of finite type. Then G has a systolic Garside presentation if and only if G = ( * p i=1 G n i ,m i )/(∆ n i ,m i = ∆ n j ,m j ∀i, j) for some positive integers p, n 1 , . . . , n p and m 1 , . . . , m p .

Proof. We start by showing that if G = ( * p i=1 G n i ,m i )/(∆ n i ,m i = ∆ n j ,m j ∀i, j) for some positive integers p, n 1 , . . . , n p , m 1 , . . . , m p the group G has a systolic Garside presentation. We start with the case p = 1. If n = m = 1, we have S = {x 1 } so the Garside presentation is systolic. Otherwise G = G n,m for some positive integers n and m, 2 ≤ m and the Garside element is ∆ = ∆ n,m = prod(x 1 , . . . , x n ; m). For simpler notation we always consider the index i modulo n. The set of simple elements is

S = {prod(x i , . . . , x i+n ; k) | 1 ≤ i ≤ n and 0 ≤ k ≤ m}.
Then for 0 ≤ k ≤ l ≤ m, we have prod(x i , . . . , x i+n ; k) ∧ L prod(x j , . . . , x j+n ; l) = e if i = j and l < m prod(x i , . . . , x i+n ; k) if i = j or l = m.

Similarly for 0 ≤ k ≤ l ≤ m, we have prod(x i , . . . , x i+n ; k) ∧ R prod(x j , . . . , x j+n ; l)

= e if i + k ≡ j + l and l < m prod(x i , . . . , x i+n ; k) if i + k ≡ j + l or l = m.
So by Theorem 4.4 the Garside presentation of G n,m is systolic. Now consider the case p > 1. Then the element ∆ = ∆ n 1 ,m 1 = • • • = ∆ np,mp is the Garside element of G. Let S be the set of simple elements of G and S i the set of simple elements of G n i ,m i for i = 1, . . . , p. Then S = p i=1 (S i \ {∆ n i ,m i , e}) {∆, e} is a partition of the set of simple elements. For every i, j ∈ {1, . . . , p}, S i satisfies s ∧ L t, s ∧ R t ∈ {e, s, t} for all s, t ∈ S i and we have s ∧ R t = s ∧ L t = e if s ∈ S i \ {∆}, t ∈ S j \ {∆}, i = j. So for all s, t ∈ S we have s ∧ R t, s ∧ L t ∈ {s, t, e}. Hence by Theorem 4.4, G has a systolic Garside presentation. Now assume G is a Garside group with systolic Garside presentation. Let ∆ be the Garside element, P the monoid of positive words, S the set of simple elements and A the set of atoms of P . So by Theorem 4.4, we have for all s, t ∈ S, s ∧ L t, s ∧ R t ∈ {e, s, t}. First note that if ∆ ∈ A we have S = {e, ∆}, A = {∆}, G = Z and P = N. Hence we can write G as ( * p i=1 G n i ,m i )/(∆ n i ,m i = ∆ n j ,m j ∀i, j) with p = 1 and n 1 = m 1 = 1. So we can now assume that ∆ / ∈ A. We start with showing that for all a ∈ A there exists a unique ξ(a) ∈ A such that aξ(a) ∈ S. As ∆ / ∈ A, such a ξ(a) ∈ A exists. Assume it is not unique, so let a 1 , a 2 ∈ A with aa 1 , aa 2 ∈ S. Since aa 1 ∧ L aa 2 ∈ {e, aa 1 , aa 2 } and a, a 1 , a 2 are atoms, aa 1 = aa 2 and hence a 1 = a 2 . Similarly for all a ∈ A there exists a unique ρ(a) ∈ A such that ρ(a)a ∈ S. In particular for all a ∈ A, ρ(ξ(a)) = a = ξ(ρ(a)). So the map ξ : A → A is bijective with inverse map ρ : A → A. As A is finite, the map ξ is a permutation of A. So the orbits of ξ form a partition of A and ξ can be written as a product of cycles of disjoint support, ξ = (a 1,1 , a 1,2 , . . . , a 1,n 1 )(a 2,1 , . . . , a 2,n 2 ) • • • (a p,1 , . . . , a p,np ). Then for any 1 ≤ i ≤ p there exists m i ≥ 2 such that ∆ = prod(a i,1 , . . . , a i,n i ; m i ). Since ∆ is balanced we have prod(a i,1 , . . . , a i,n i ; m i ) = prod(a i,2 , a i,3 , . . . , a i,n i , a i,1 ; m i ) = • • • = prod(a i,n i , a i,1 , . . . , a i,n i -1 ; m i ).

The set A i = {a i,1 , . . . , a i,n i } corresponds to the atoms of G n i ,m i . Hence

G = ( * p i=1 G n i ,m i )/(∆ n i ,m i = ∆ n j ,m j ∀i, j).
Corollary 4.7.

1. The group

G n,n = x 1 , . . . , x n | x 1 x 2 . . . x n = x 2 x 3 . . . x n x 1 = • • • = x n x 1 . . . x n-1 ,
which is the fundamental group of the complement of n lines through the origin in C 2 , has a systolic Garside presentation with respect to the Garside element ∆ = x 1 x 2 . . . x n .

2. Consider n positive integers p 1 , . . . , p n , p i ≥ 2. The group

G = x 1 , . . . , x n | x p 1 1 = x p 2 2 = • • • = x pn n
has a systolic Garside presentation with respect to the Garside element ∆ = x p 1 1 . In particular torus knot groups x, y | x p = y q have a systolic Garside presentation with respect to the Garside element ∆ = x p . Remark 4.8. In [START_REF] Dehornoy | Gaussian groups and Garside groups, two generalisations of Artin groups[END_REF], Example 5 mentions a generalization of the groups G Remark 4.9 (Restrictions on systolicity in Garside groups). Consider a Garside group G with Garside element ∆. Then ∆ k is in the center of G for some positive integer k. Let S be the set of simple elements. Suppose there is some balanced element

n,m . Let p, n, m ∈ N, 2 ≤ m, 2 ≤ p ≤ n. It claims that K n,p,m = x 1 , x 2 , . . . , x n | prod(x 1 , . . . , x p ; m) = prod(x 2 , . . . , x p+1 ; m) = . . . = prod(x n-p+1 , . . . , x n ; m) = prod(x n-p+2 , . . . , x n , x 1 ; m) = . . . = prod(x n , x 1 , . . . , x p-1 ; m)
δ ∈ S \ {∆, e}. Let T = {a ∈ G | 1 ≤ L a ≤ L δ} = {a ∈ G | 1 ≤ R a ≤ R δ}.
Let a ∈ T be an atom and suppose δ / ∈ a . Then δ l is in the center of the subgroup of G generated by T for some positive integer l. If T = G, we have a, δ l , ∆ k ∼ = Z 3 . This implies in particular that G is not systolic. Given a finite labeled simple graph Γ, the Artin group associated to Γ is given by The Artin monoid associated to Γ is given by Recall the following definitions: An orientation on a simple graph Γ is an assignement o(e) for each edge e ∈ E(Γ) where o(e) is a set of one or two endpoints of e. An edge with both endpoints assigned is bioriented. The startpoint i(e) is an assignement of one or two startpoints of e, which is consistent with the choice of o(e). If o(e) consists of one point, i(e) consists of one point such that e = (i(e), o(e)), if o(e) consists of two points then so does i(e). We say that a cycle γ is directed if for each v ∈ γ there is exactly one edge e ∈ γ with v ∈ o(e). A cycle is undirected if it is not directed. We say that a 4-cycle γ

Artin groups

A Γ = s v , v ∈ V | [s v s w s v . . . ] me = [s w s v s w . . .
A + Γ = s v , v ∈ V | [s v s w s v .
= (a 1 , a 2 , a 3 , a 4 ) is misdirected if a 2 ∈ o(a 1 , a 2 ), a 2 ∈ o(a 2 , a 3 ), a 4 ∈ o(a 3 , a 4 ) and a 4 ∈ o(a 4 , a 1 ). a 1 a 4 a 3 a 2
Lemma 4.11. Let Γ be a labeled simple graph with an orientation o such that an edge is bioriented if and only if it has label 2. Assume that every 3-cycle in Γ is directed.

Let V (Γ) = {v 1 , . . . , v n }. Consider the set S = {x 1 , x 2 , . . . , x n } ∪ {∆ e , t e 1 , t e 2 , . . . , t e me-2 | e ∈ E(Γ) with label m e }. For each e ∈ E(Γ) with label m e ≥ 3 and with i(e) = v i and o(e) = v j , we consider the set R e = {x i x j ∆ -1 e , x j t e 1 ∆ -1 e , t e 1 t e 2 ∆ -1 e , . . . , t e me-3 t e me-2 ∆ -1 e , t e me-2 x i ∆ -1 e }. For each e ∈ E(Γ) with label m e = 2 and with o(e) = i(e) = {v i , v j } we consider the set R e = {x i x j ∆ -1 e , x j x i ∆ -1 e }. Let R = e∈E(Γ) R e . Then the presentation S | R is a restricted triangular presentation of A Γ . We call this the dual presentation of A Γ with orientation o.

Proof. The standard presentation of A Γ is A Γ = x 1 , . . . , x n | [x i x j x i . . . ] me = [x j x i x j . . . ] me for all edges e = (v i , v j ) with label m e .

We first see that the dual presentation is indeed a presentation of A Γ . Let e ∈ E(Γ) with m e = 2, i(e) = o(e) = {v i , v j }. Then the standard presentation states x i x j = x j x i . In the dual presentation the relations R e imply ∆ e = x i x j and ∆ e = x j x i and hence x i x j = x j x i . Let e ∈ E(Γ) with m e ≥ 3, i(e) = v i and o(e) = v j . For k ∈ {1, . . . , m e -2}, the relations R e imply on one hand t e k = ([. . . x j x i x j ] k ) -1 [. . . x j x i x j ] k+1 and on the other hand t e k = [x i x j x i . . . ] me-k ([x i x j x i . . . ] me-k-1 ) -1 . This implies the relation [x i x j x i . . . ] me = [x j x i x j . . . ] me . Conversely [x i x j x i . . . ] me = [x j x i x j . . . ] me implies x i x j = [x j x i x j . . . ] me ([x i x j x i . . . ] me-2 ) -1 = x j t e 1 . For k ∈ {1, . . . , m e -3} we have t e k t e k+1 = ([. . . x j x i x j ] k ) -1 [. . . x j x i x j ] k+1 ([. . . x j x i x j ] k+1 ) -1 [. . . x j x i x j ] k+2 = ([. . . x j x i x j ] k ) -1 [. . . x j x i x j ] k+2 = x i x j .

Finally when k = m e -2 we have t e me-2 x i = ([. . . x j x i x j ] me-2 ) -1 [. . . x j x i x j ] me-1 x i = ([. . . x j x i x j ] me-2 ) -1 [. . . x i x j x i ] me = ([. . . x j x i x j ] me-2 ) -1 [. . . x j x i x j ] me = x i x j .

Let a, b, c ∈ S. To see that the dual presentation is a restricted triangular presentation we check that abc = e, abc / ∈ S and abc -1 = e ⇒ abc -1 ∈ R. Note that abc / ∈ S implies S ∩ S -1 = ∅. We consider the following map ξ : S → Z defined by ξ(∆ e ) = 2 and ξ(t e i ) = 1 for all e ∈ E(Γ), i ∈ {1, . . . , m e -2} and ξ(x i ) = 1 for i ∈ {1, . . . , n}. As ξ(a) + ξ(b) -ξ(c) = 0 for all abc -1 ∈ R, the map extends to a homomorphism ξ : A Γ → Z. For any a, b, c ∈ S, we have ξ(a) + ξ(b) ≥ 2 and ξ(abc) = ξ(a) + ξ(b) + ξ(c) ≥ 3 so it follows that ab = e, abc = e and abc / ∈ S. Also for any a, b, c ∈ S such that abc -1 = e we have ξ(abc -1 ) = 0, which implies ξ(c) = ξ(a) + ξ(b) so we necessarily have c = ∆ e for some e ∈ E(Γ) and a, b ∈ S \ {∆ e , e ∈ E(Γ)}. So fix e ∈ E(Γ) and c = ∆ e = x i x j . Let S 0 = S \ {∆ e , e ∈ E(Γ)}. We need to verify for all (a, b) ∈ S 0 × S 0 that if abc -1 = e we have abc -1 ∈ R.

Our proof relies on the following property of A + Γ : Let α = x i 1 x i 2 . . . x i l and β = x j 1 x j 2 . . . x j k be two words on x 1 , . . . , x n . They represent the same element in A + Γ if and only if we can transform α into β using a finite number of transformations of the form u•[x p x q x p . . . ] m f •u = u•[x q x p x q . . . ] m f •u for some f = (v p , v q ) ∈ E(Γ). If α does not contain a subword of this form, no transformation is possible and the expression is unique.

Let a = x k and b = x l for some k, l ∈ {1, . . . , n}. Then ab = c implies x k x l = x i x j in A Γ hence the equality also holds in A + Γ . If m e ≥ 3 this necessarily

3)

  If ∃u, v, x, b, c ∈ S, v = x, with ux ∈ S, uv ∈ S and vb = xc ∈ S then ∃k ∈ S such that k = uvb or v = xk. 4) If ∃v, w, x, a, d ∈ S, v = x, with vw ∈ S, xw ∈ S and dx = av ∈ S then ∃k ∈ S such that k = avw or x = kv. 5) If ∃u, v, w, x ∈ S, v = x, u = w, with wv ∈ S, wx ∈ S, uv ∈ S and ux ∈ S then ∃k ∈ S such that w = ku or x = vk.

  m. and prod(x 1 , . . . , x p ; 0) = e. Consider the group

  Consider the notation [xyx . . . ] k = xyx . . .k and [. . . xyx] k = . . . xyx k for some k ∈ N. Given a finite labeled simple graph Γ, the Artin group associated to Γ is given by

  and edges e[g, a], e[ga, b], e[gab, c] or e[g, a], e[ga, b], e[g, c]. On the other hand each 2-simplex has vertices and edges which correspond to such a triple of generators. This implies that π 1 (F lag(G, S)/G) = S | a • b • c = e or a • b • c -1 = e for all a, b, c

  Lemma 3.4. The link L is 6-large if and only if the following additional conditions on S are satisfied: 1) If ∃u, w, a, b, c, d ∈ S, u = w, a = d, with ua = wb ∈ S and ud = wc ∈ S then ∃k ∈ S such that w = uk or ua = udk. 2) If ∃v, x, a, b, c, d ∈ S, v = x, a = b, with bv = cx ∈ S and av = dx ∈ S then ∃k ∈ S such that v = kx or av = kbv. 3) If ∃u, v, x, b, c ∈ S, v = x, with ux ∈ S, uv ∈ S and vb = xc ∈ S then ∃k ∈ S such that k = uvb or v = xk. 4) If ∃v, w, x, a, d ∈ S, v = x, with vw ∈ S, xw ∈ S and dx = av ∈ S then ∃k ∈ S such that k = avw or x = kv. 5) If ∃u, v, w, x ∈ S, v = x, u = w, with wv ∈ S, wx ∈ S, uv ∈ S and ux ∈ S then ∃k ∈ S such that w = ku or x = vk.

1)

  If ∃u, w, a, b, c, d ∈ S, u = w, a = d, with ua = wb ∈ S and ud = wc ∈ S then ∃k ∈ S such that w = uk or ua = udk. 2) If ∃v, x, a, b, c, d ∈ S, v = x, a = b, with bv = cx ∈ S and av = dx ∈ S then ∃k ∈ S such that v = kx or av = kbv. 3) If ∃u, v, x, b, c ∈ S, v = x, with ux ∈ S, uv ∈ S and vb = xc ∈ S then ∃k ∈ S such that k = uvb or v = xk. 4) If ∃v, w, x, a, d ∈ S, v = x, with vw ∈ S, xw ∈ S and dx = av ∈ S then ∃k ∈ S such that k = avw or x = kv. 5) If ∃u, v, w, x ∈ S, v = x, u = w, with wv ∈ S, wx ∈ S, uv ∈ S and ux ∈ S then ∃k ∈ S such that w = ku or x = vk.

4)

  If ∃u, v, w, x ∈ S pairwise distinct with vw ∈ S and xw ∈ S and x ≤ R u and v ≤ R u then uw ∈ S or v ≤ R x. 5) If ∃u, v, w, x ∈ S pairwise distinct with uv ∈ S and ux ∈ S and wv ∈ S and wx ∈ S then v ≤ L x or w ≤ R u.

3 )

 3 If ∃u, v, x, b, c ∈ S, v = x with ux ∈ S, uv ∈ S and vb = xc ∈ S then u ≤ L ux and u ≤ L uv so uv ∧ L ux = e, so either uv ∧ L ux = uv or uv ∧ L ux = ux, say uv ∧ L ux = ux so uv = uxk for some k ∈ S and then v = xk. 4) If ∃v, w, x, a, d ∈ S, v = x with vw ∈ S, xw ∈ S and dx = av ∈ S then w ≤ R vw and w ≤ R xw so vw ∧ R xw = e, so either vw ∧ R xw = vw or vw ∧ R xw = xw, say xw ∧ R vw = vw so xw = kvw for some k ∈ S and then x = kv. 5) If ∃u, v, w, x ∈ S, v = x, u = w with wv ∈ S, wx ∈ S, uv ∈ S and ux ∈ S then u ≤ L uv and u ≤ L ux so uv ∧ L ux = e, so either uv ∧ L ux = uv or uv ∧ L ux = ux, say uv ∧ L ux = uv so uv = uxk for some k ∈ S and then v = xk.

  First assume ∃a, b ∈ S, a = b, with a ∧ L b = c for some c ∈ S \ {a, b}, i.e c = e, a, b. Then ∃k a , k b , r a , r b ∈ S with a = ck a , b = ck b and ∆ = k a r a = k b r b . Then c, k a , k b , r a , r b ∈ S and k a = k b , ck a = a ∈ S, ck b = b ∈ S and ∆ = k a r a = k b r b ∈ S. But ck a r a = c∆ / ∈ S and k ∈ S with k a = k b k since c = a ∧ L b = b, similarly k ∈ S with k b = k a k since c = a ∧ L b = a. So condition 3) of Theorem 3.5 fails. So F lag(G, S) is not systolic. Finally assume ∃a, b ∈ S, a = b, with a ∧ R b = c for some c ∈ S \ {a, b}, i.e. c = e, a, b. Then ∃k a , k b , r a , r b ∈ S with a = k a c, b = k b c and ∆

  is a Garside group by [DP99] Proposition 5.2. But this is a wrong application of [DP99] Proposition 5.2, as one can see by considering for example the case n = 5, p = 3 and m = 4 or more generally m = p + 1. So the question of whether K n,p,m is a Garside group when p = n remains open. Question: For k ≥ 2, the group G k = a, b | b k = aba is Garside with Garside element ∆ = b k+1 = abab = baba. The elements b k and bab are both simple elements, but b k ∧ L bab = b / ∈ {e, b k , bab}. So it does not satisfy the conditions of Theorem 4.4. Is it systolic? Is there another Garside structure on this group for which it has a systolic Garside presentation?

  Recall the notation [xyx . . . ] k = xyx . . . k and [. . . xyx] k = . . . xyx k for some k ∈ N.

  ] me for all edges e = (v, w) with label m e . For n ∈ N ≥2 , the dihedral Artin group DA n is the Artin group defined by the graph a n b . So DA n = a, b | [aba . . . ] n = [bab . . . ] n .

  . . ] me = [s w s v s w . . . ] me for all edges e = (v, w) with label m e + . By [Par02] Theorem 1.1, the canonical homomorphism ι : A + Γ → A Γ is an injection. Corollary 4.10. The dihedral Artin group DA n is Cayley systolic for all n ∈ N ≥2 . Proof. The dihedral Artin group DA n corresponds to the Garside group G 2,n with Garside element ∆ = [aba . . . ] n . So by Theorem 4.6 it is Cayley systolic.

Every finitely presented subgroup of G is systolic. Wise showed this for finitely presented subgroups of torsion-free subgroups. For all systolic groups, this has been shown in[START_REF] Zadnik | Finitely presented subgroups of systolic groups are systolic[END_REF],[START_REF] Gaelan | Lifting group actions, equivariant towers and subgroups of non-positively curved groups[END_REF].

Acknowledgements The author thanks their advisors Thomas Haettel and Luis Paris for many helpful discussions and advice. The author is partially supported by the French project "AlMaRe" (ANR-19-CE40-0001-01) of the ANR.

implies k = i and l = j. If m e = 2, this implies either k = i and l = j or k = j and l = i. In all of these cases the corresponding relation is in R.

Let a = x k and b = t f l with k ∈ {1, . . . , n}, f = (v p , v q ) ∈ E(Γ), m f ≥ 3 and 1 ≤ l ≤ m f -2. Set l = m f -l -1, so 1 ≤ l ≤ m f -2. Then ab = c implies x k t f l = x i x j so x k [x p x q x p . . . ] m f -l ([x p x q x p . . . ] m f -l-1 ) -1 = x i x j and hence x k [x p x q x p . . . ] l +1 = x i x j [x p x q x p . . . ] l in A + Γ . As all 3-cycles are directed, x i , x j , x k cannot commute with both x p and x q . So the last letter on the left hand side and on the right hand side is different and is either x p or x q . The only way to change this last letter is to apply the relation [x p x q x p . . . ] m f = [x q x p x q . . . ] m f . So we need k = q and l = m f -2 so l = 1. This implies the equality x p x q = x i x j so p = i and j = q. Hence f = e, a = x j , b = t e 1 and the corresponding relation is in R.

Then ab = c implies t f l x k = x i x j so ([. . . x q x p x q ] l ) -1 [. . . x q x p x q ] l+1 x k = x i x j so [. . . x q x p x q ] l+1 x k = [. . . x q x p x q ] l x i x j in A + Γ . As all 3-cycles are directed, x i , x j , x k cannot commute with both x p and x q . The first letter on the left hand side and on the right hand side is different and is either x p or x q . The only way to change this first letter is to use the relation [x p x q x p . . . ] m f = [x q x p x q . . . ] m f . So we need k = p and l = m f -2. This implies x p x q = x i x j hence i = p and j = q. So f = e, a = t e me-2 and b = x i and the corresponding relation is in R.

Finally let

Γ . As all 3-cycles are directed, x r and x i cannot commute with both x p and x q . Similarly x q and x j cannot commute with both x r and x s . So the first letter on the left and on the right hand side is different and is either x p or x q . Similarly the last letter on the left and on the right hand side is different and is either x r or x s . So we need to apply the relations [x p x q x p . . . ] m f = [x q x p x q . . . ] m f and [x r x s x r . . . ] mg = [x s x r x s . . . ] mg . This requires r = p and s = q and m f ≤ k + l + 2. If k + l + 2 > m f , we can apply the relation [x p x q x p . . . ] m f = [x q x p x q . . . ] m f on a piece of the left hand side of length m f . But since k + l + 2 < 2m f , this only allows us to change either the first or the last letter but not both. So we necessarily have k

Applying the relation leads to x i x j = x p x q and i = p and j = q. So f = g = e, a = t e k and b = t e k+1 for some k ∈ {1, . . . , m e -3} and the corresponding relation is in R.

So we have indeed abc = 1, abc / ∈ S and abc -1 = e ⇒ abc -1 ∈ R for a, b, c ∈ S so this is a restricted triangular presentation.

Remark 4.12. Let Γ be a simple graph, with edges labeled by numbers ≥ 2 and with an orientation o such that an edge is bioriented if and only if it has label 2.

Then it follows from the proof of Lemma 4.11 that the dual presentation of A Γ with orientation o is always a presentation of A Γ but it is not always a restricted triangular presentation. For example if Γ is a 3-cycle with one edge labeled by 2 and the two others labeled by m, n ≥ 3, the presentation is not a restricted triangular presentation.

Remark 4.13. In Corollary 4.10 we used the Garside structure on DA n induced by G 2,n . Lemma 4.11 implies that DA n ∼ = G n,2 , this presentation corresponds to another Garside structure. So in particular G 2,n ∼ = G n,2 as groups but with different Garside structures.

Theorem 4.14. Let Γ be a simple graph, with edges labeled by numbers ≥ 2 and with an orientation o such that an edge is bioriented if and only if it has label 2. Assume that every 3-cycle is directed and no 4-cycle is misdirected. Let A be the Artin group associated to Γ. Then A is Cayley systolic.

Proof. We prove using Theorem 3.5 that F lag(A Γ , S) with respect to the dual presentation with orientation o is systolic. Note that with respect to the partial order defined in Remark 3.8:

• If a, b ∈ S with ab ∈ S then ab = ∆ e for some e ∈ E(Γ).

• If s ≤ L ∆ e and s ≤ L ∆ f for some s ∈ S, e, f ∈ E(Γ), e = f then s ∈ {x 1 , x 2 , . . . , x n }.

• If s, t ≤ L ∆ e and s, t ≤ L ∆ f for some s, t ∈ S, s = t then e = f .

• If ∆ e = x i x j , i(e) ∪ o(e) = {x i , x j }.

We check the different conditions of Theorem 3.5: 1) Assume ∃u, w, a, b, c, d ∈ S, u = w, a = d with ua = wb ∈ S and ud = wc ∈ S. Then ua = wb = ∆ e and ud = wc = ∆ f for some e, f ∈ E(Γ). As u = w we have e = f . But then a = d which is a contradiction.

2)Assume ∃v, x, a, b, c, d ∈ S, v = x, a = b with bv = cx ∈ S and av = dx ∈ S. Then bv = cx = ∆ e and av = dx = ∆ f for some e, f ∈ E(Γ). As v = x we have e = f . And so a = b which is a contradiction.

3) Assume ∃u, v, x, b, c ∈ S, v = x with ux, uv, vb, xc ∈ S and vb = xc. Then uv = ∆ e , ux = ∆ f and vb = xc = ∆ g for some e, f, g ∈ E(Γ).

• If e = f we have x = v which is a contradiction.

• If e = g and e = f then u, x ∈ {x 1 , . . . , x n }.

But then o(e) ∪ i(e) = {u, x} = o(f ) ∪ i(f ). So e = f , which is a contradiction.

• If e = g, f = g, e = f then u, v, x ∈ {x 1 , . . . , x n } and u ∈ i(e), u ∈ i(f ), v ∈ o(e), x ∈ o(f ), and {x, v} = i(g) ∪ o(g). But this corresponds to an undirected triangle in the defining graph Γ.

4) Assume ∃v, w, x, a, b ∈ S, v = x, with vw, xw, dx, av ∈ S and dx = av. Then vw = ∆ e , xw = ∆ f and dx = av = ∆ g for some e, f, g ∈ E(Γ).

• If e = f we have x = v which is a contradiction.

• If e = g and e = f then w, x ∈ {x 1 , . . . , x n }.

But then o(e) ∪ i(e) = {w, x} = o(f ) ∪ i(f ). So e = f , which is a contradiction.

• If e = g, f = g, e = f then v, w, x ∈ {x 1 , . . . , x n } and w ∈ o(e), w ∈ o(f ) and v ∈ i(e), x ∈ i(f ) and o(g) ∪ i(g) = {v, x}. But this corresponds to an undirected triangle in the defining graph Γ.

5) Assume ∃u, v, w, x ∈ S, v = x, u = w with wv, wx, uv, ux ∈ S. Then wv = ∆ e , wx = ∆ f , uv = ∆ g and ux = ∆ h for some e, f, g, h ∈ E(Γ). Then v = x implies e = f and g = h, and u = w implies e = g and f = h. So u, v, w, x ∈ {x 1 , . . . , x n } and v ∈ o(e) ∩ o(g), x ∈ o(f ) ∩ o(h), w ∈ i(e) ∩ i(f ) and u ∈ i(g) ∩ i(h). Furthermore i(e) ∪ o(e) = {w, v} which implies v = w and i(h) ∪ o(h) = {u, x} which implies u = x. So the 4-cycle (u, v, w, x) is misdirected. This is a contradiction to the orientation on Γ.