# New Garside structures and applications to Artin groups

Thomas Haettel and Jingyin Huang

Abstract. Garside groups are combinatorial generalizations of braid groups
which enjoy many nice algebraic, geometric, and algorithmic properties. In
this article we propose a method for turning the direct product of a group $G$
by $\mathbb{Z}$ into a Garside group, under simple assumptions on $G$. This
method gives many new examples of Garside groups, including groups satisfying
certain small cancellation condition (including surface groups) and groups
with a systolic presentation.

Our method also works for a large class of Artin groups, leading to many new
group theoretic, geometric and topological consequences for them. In
particular, we prove new cases of $K(\pi,1)$-conjecture for some hyperbolic
type Artin groups.

††footnotetext: Thomas Haettel<EMAIL_ADDRESS>IMAG, Univ
Montpellier, CNRS, France, and IRL 3457, CRM-CNRS, Université de Montréal,
Canada.
Jingyin Huang, Department of Mathematics, The Ohio State University, 100 Math
Tower, 231 W 18th Ave, Columbus, OH 43210, U.S.
Keywords : Garside groups, Artin groups, $K(\pi,1)$-conjecture, hyperplane
arrangements, Coxeter groups, dual Garside structure, combinatorial Garside
structure, interval groups, CUB spaces, Helly graphs, Helly groups, bicombing,
Nonpositive curvature AMS codes : 20E42, 20F36, 20F55, 05B35, 06A12, 20F65,
05C25

##  1 Introduction

The notion of Garside group originated in Garside’s work on word problems and
conjugacy problems for braid groups [Gar69]. It turns out the key structure
needed in Garside’s argument also appears in more general groups later,
notably in spherical Artin groups [BS72] and fundamental groups of
complexified central simplicial arrangement complements [Del72]. An axiomatic
setting up was provided in [DP99, Deh02], to study groups that share a similar
structure as a class, called _Garside groups_. Since then, other important
classes of groups were proven to be Garside groups, including but not limited
to some semi-direct products [CP05], some complex braid groups [Bes15, CP11,
CLL15], structure groups of non-degenerate, involutive and braided set-
theoretical solutions of the quantum Yang-Baxter equation [Cho10],
crystallographic braid groups [MS17] etc. Garside groups are also known to be
closed under certain kind of amalgamation products and HNN extensions [Pic22],
as well as Zappa–Szép products [GT16]. There are also a number of variations
and generalizations of Garside groups, applying to more natural examples - we
refer to the book [Deh15] for a comprehensive review.

Garside groups in this article always means Garside groups of finite type,
i.e. the Garside element has finitely many divisors. If it has infinitely
many, then we will call it a _quasi-Garside_ group. Garside groups are known
to enjoy a long list of nice geometric, group theoretic and topological
properties - they are biautomatic [Cha92a, Deh02], hence have solvable word
problems and conjugacy problems, they are torsion-free, and admit finite
$K(\pi,1)$ spaces [CMW04, DL03], they act geometrically on Helly graphs and on
injective metric spaces [HO21b, Hae21], hence satisfies the Farrell-Jones
conjecture and coarse Baum-Connes conjecture [CCG+20] etc. Also Garside groups
of finite or infinite type plays central role in the proof of the
$K(\pi,1)$-conjecture for different classes of complex hyperplane complements,
see [Del72, Bes15, MS17, PS21].

###  1.1 New Garside groups

While Garside groups enjoy nice properties, they have a very strong algebraic
constraint: since a power of the Garside element is central, they have
infinite center. This explains why the list of known examples of Garside
groups is somehow limited. In this article, we propose a simple approach to
circumvent this obstruction and use Garside theory to study some groups with
possibly trivial center. Namely, given a group $G$, we will consider the
direct product of $G$ with $\mathbb{Z}$ to artificially create a center, which
will serve as the Garside axis. Then we work backward to find necessary
conditions on $G$ to make sure $G\times\mathbb{Z}$ is actually a Garside
group, leading to the following simple criterion.

######  Theorem A.

(=Theorem 3.9) Let $U$ be a finite set, endowed with a positive partial
multiplication (see Definition 3.1), and associated prefix order
$\leqslant_{L}$ and suffix order $\leqslant_{R}$. Assume that the following
hold:

  * $\bullet$

$(U,\leqslant_{L})$ and $(U,\leqslant_{R})$ are semilattices.

  * $\bullet$

For any $a,u,v,w\in U$ such that $a\cdot u,a\cdot v\in U$ and $w$ is the join
for $\leqslant_{L}$ of $u$ and $v$, then $a\cdot w\in U$.

  * $\bullet$

For any $a,u,v,w\in U$ such that $u\cdot a,v\cdot a\in U$ and $w$ is the join
for $\leqslant_{R}$ of $u$ and $v$, then $w\cdot a\in U$.

  * $\bullet$

For any $a,b,u,v\in U$ such that $a\cdot u,a\cdot v,b\cdot u,b\cdot v\in U$,
either $a,b$ have a join for $\leqslant_{R}$, or $u,v$ have a join for
$\leqslant_{L}$.

Consider the group $G_{U}$ given by the following presentation:

$G_{U}=\langle U\,|\,\forall u,v,w\in U\mbox{ such that }u\cdot v=w,\mbox{ we
have }uv=w\rangle.$

Then the group $G_{U}\times\mathbb{Z}$ is a Garside group, with Garside
element $(e,1)$.

First note that if a group $G$ is such that $G\times\mathbb{Z}$ is a Garside
group, then we can deduce an impressive list of consequences for $G$, see
Theorem D below.

This method can be applied to several classes of groups that we discuss in
this article. For instance, it applies to some groups given by a $T(5)$
positive presentation, see Theorem 3.10 for a precise statement. In
particular, we deduce the following nice consequence.

######  Corollary B.

(=Corollary 3.11) For any surface $S$ of finite type (possibly non-
orientable), except the projective plane, $\pi_{1}(S)\times\mathbb{Z}$ is a
Garside group.

Another interesting family of examples comes from groups given by a
presentation such that the associated flag Cayley complex is systolic, called
_systolic restricted presentation_ by Soergel in [Soe21], where they are
defined and studied. We refer to Definition 3.12. Examples include some
amalgams of Garside groups and some $2$-dimensional Artin groups. For these
groups, we prove the following.

######  Corollary C.

(=Corollary 3.13) Let $G$ denote a group with a systolic restricted
presentation. Then $G\times\mathbb{Z}$ is a Garside group.

Theorem A also applies a class of groups with positive square presentations in
the sense of Definition 3.14, where a criterion for such groups times
$\mathbb{Z}$ to be Garside is provided in Theorem 3.15. This applies to a
subclass of groups arising from word labeled oriented graphs in the sense of
[HR15], as well as some of the mock right-angled Artin groups defined in
[Sco08].

In order to motivate the study of groups $G$ for which $G\times\mathbb{Z}$ is
a Garside group, we record here a list of direct consequences. We recall the
definition of Garside groups in Section 2.4, and we recall various nonpositive
curvature notions in Section 2.5.

######  Theorem D.

Assume that $G$ is a group such that $G\times\mathbb{Z}$ is Garside. Then the
following hold:

  1. 1.

The group $G\times\mathbb{Z}$ is Helly.

  2. 2.

The group $G$ is torsion-free.

  3. 3.

The group $G$ is CUB, more precisely it acts geometrically on a finite-
dimensional metric space with a unique convex geodesic bicombing. Moreover,
this metric space is a simplicial complex such that each simplex is equipped
with a polyhedral norm.

  4. 4.

The group $G$ acts geometrically on a weakly modular graph.

  5. 5.

The group $G$ is biautomatic, and in particular:

     * $\bullet$

The centralizer of a finite set of elements of $G$ is biautomatic.

     * $\bullet$

$G$ has solvable word and conjugacy problems.

     * $\bullet$

Any polycyclic subgroup of $G$ is virtually abelian, finitely generated and
undistorted.

     * $\bullet$

$G$ has quadratic Dehn function, as well as Euclidean higher dimensional Dehn
function.

  6. 6.

Any element of $G$ has rational translation length, with uniformly bounded
denominator.

  7. 7.

The group $G$ has contractible asymptotic cones.

  8. 8.

The group $G$ satisfies the Farrell-Jones conjecture with finite wreath
products.

  9. 9.

The group $G$ satisfies the coarse Baum-Connes conjecture.

  10. 10.

The group ring $\mathbb{K}[G]$ satisfies Kaplansky’s idempotent conjecture, if
$\mathbb{K}$ is a field with characteristic zero.

We defer the references for this theorem to Section 2.1.

###  1.2 Applications to Artin groups

One of the main motivation for our work comes from Artin groups, see Section
2.2 for basic definitions. To each Coxeter group, there is an associated Artin
group, in the same fashion that the $n$-strand braid group is associated to
the symmetric group of order $n$. General Artin groups are largely mysterious,
and even basic questions such as the following are still widely open (see
[GP12], [Cha], [McC17]).

  1. 1.

Are Artin groups torsion-free?

  2. 2.

What is the center of Artin groups?

  3. 3.

Do Artin groups have solvable word problem?

  4. 4.

Is the natural hyperplane complement a classifying space for Artin groups (the
$K(\pi,1)$ conjecture, see Section 2.3)?

Note that a positive answer to the $K(\pi,1)$ conjecture implies that the
corresponding Artin group is torsion-free, and also that its center is known
(see [JS23]).

For Artin groups of spherical type, i.e. when the associated Coxeter group is
finite, all these questions have a precise answer, which all rely on the
existence of Garside structures. In fact, Artin groups of spherical type enjoy
two different Garside structures: the standard one, associated with the
longest element in the associated finite Coxeter group, and the dual one,
associated with a Coxeter element. For an Artin group of non-spherical type,
only the dual structure could be studied. In this case, the dual interval is
always infinite, so one can only hope for a quasi-Garside structure, which has
much fewer consequences. Nevertheless, it is known that for an Artin group of
affine type $\widetilde{A_{n}}$, $\widetilde{C_{n}}$ or $\widetilde{G_{2}}$
([Dig06, Dig12, McC15]), or for an Artin group of rank $3$ [DPS22], this dual
structure turns the Artin group into a quasi-Garside group. In fact, for every
Artin group of affine type, McCammond and Sulway manage to provide a natural
embedding of the Artin group into a quasi-Garside crystallographic braid
group, which is central in the proof of the $K(\pi,1)$ conjecture by Paolini
and Salvetti ([PS21]).

However, even though a quasi-Garside structure might be sufficient to find
classifying spaces, we already mentioned that a Garside structure on the
direct product with $\mathbb{Z}$ is much more interesting, see Theorem D. In
order to state our results concerning Artin groups, let us first recall some
notations, we refer to Section 2.2 for more details on our notations on Artin
groups and their associated Coxeter groups. In particular, each Artin group or
Coxeter group has a Coxeter presentation graph $\Gamma$, and a Dynkin diagram
$\Lambda$. We will write $A_{\Gamma}$ (resp. $W_{\Gamma}$) to denote the Artin
group (resp. Coxeter group) with Coxeter presentation graph $\Gamma$.

We say an Artin group is of _cyclic type_ if its Dynkin diagram is a cycle,
and any proper parabolic subgroup is spherical. We refer to Table 1 for a
complete list of cyclic type Artin groups. In particular, it contains some
Artin groups that are associated with certain Coxeter groups acting on the
hyperbolic spaces $\mathbb{H}^{3}$ or $\mathbb{H}^{4}$ \- all of the four
basic questions are open for these Artin groups.

######  Theorem E.

(=Proposition 5.6) Suppose $A_{\Gamma}$ is of cyclic type. Then
$A_{\Gamma}\times\mathbb{Z}$ is a Garside group.

As we will see later (Corollary H), Theorem E gives rise to new examples of
Artin groups satisfying the $K(\pi,1)$-conjecture. We emphasize that an
advantage of the method here is that it not only gives the
$K(\pi,1)$-conjecture, also it implies a long list of highly nontrivial
algorithmic, geometric and topological consequences as in Theorem D.

We can also treat a much more general class of Artin groups which are obtained
by gluing cyclic Artin groups and spherical Artin groups in the following way.

Given a 4-cycle $\omega\subset\Gamma$ with consecutive vertices
$\\{x_{i}\\}_{i=1}^{4}$, a pair of antipodal vertices in $\omega$ means either
the pair $\\{x_{1},x_{3}\\}$, or the pair $\\{x_{2},x_{4}\\}$. A 4-cycle in
$\Gamma$ has _diagonal_ means it has a pair of antipodal vertices of $\omega$
which are connected by an edge in $\Gamma$. We say an induced subgraph of
$\Gamma$ is of _cyclic type_ or _spherical type_ if the Artin group defined on
this subgraph is of cyclic type or spherical type. An edge of $\Gamma$ is
_large_ if it has label $\geq 3$. For an induced subgraph $\Lambda$ of
$\Gamma$, let $\Lambda^{\perp}$ be the induced subgraph of $\Gamma$ spanned by
vertices of $\Gamma\setminus\Lambda$ that commute with each vertex of
$\Lambda$.

######  Theorem F.

(=Theorem 6.3) Let $\Gamma$ be a Coxeter presentation graph such that

  * $\bullet$

each complete subgraph of $\Gamma$ is a join of a cyclic type graph and a
spherical type graph (we allow one of the join factors to be empty);

  * $\bullet$

for any cyclic type induced subgraph $\Lambda\subset\Gamma$, $\Lambda^{\perp}$
is spherical.

We assume in addition that there exists an orientation of all large edges of
$\Gamma$ such that

  1. 1.

the orientation restricted to each cyclic type subgraph of $\Gamma$ gives a
consistent orientation on the associated circle;

  2. 2.

if $\omega$ is a 4-cycle in $\Gamma$ with a pair of antipodal points $x_{1}$
and $x_{2}$ such that each edge of $\omega$ containing
$x_{i}\in\\{x_{1},x_{2}\\}$ is either not large or oriented towards $x_{i}$,
then the cycle has a diagonal.

Then $A_{\Gamma}\times\mathbb{Z}$ is a Garside group.

Below we include two simple examples of Coxeter presentation graph $\Gamma$
where the Theorem F applies, see Figure 1. The first is an amalgamation of two
Artin groups of type $\widehat{A}_{4}$ along a spherical parabolic subgroup of
type $A_{3}$. The second examples is a bit more complicated, made of a few
cyclic type Artin groups glued together in a cyclic way. Note that the edges
without label are assumed to be labeled by $2$.

$3$$3$$3$$3$$3$$3$$3$$3$ | $3$$3$$5$$4$$3$$4$$3$$4$$5$$3$$3$$3$$3$$3$$3$$3$$3$$3$$3$$3$
---|---
Figure 1: Examples of Artin groups to which Theorem F applies.

In particular, all consequences listed in Theorem D hold this class of Artin
groups. All of these consequences are new for this class, including the
solvability of word problem. As a more precise comparison to previous results,
we view the class of Artin groups in the above theorem as a combination of
basic building blocks made of cyclic type Artin groups and spherical Artin
groups. Then

  1. 1.

All consequences listed in Theorem D were known before for spherical Artin
groups [Cha92b, HO21b, HH22], hence also known for the Artin group of type
$\widetilde{A}_{n}$, as the direct product of this Artin group with
$\mathbb{Z}$ has finite index in a spherical Artin groups [KIP02];

  2. 2.

All consequences of Theorem D except the first one (acting geometrically on a
Helly graph) are known before for cyclic Artin groups with at most three
generators - as these groups act geometrically on $\operatorname{CAT}(0)$
complexes made of equilateral triangles [BM00];

  3. 3.

All consequences of Theorem D are new for the remaining cyclic type Artin
groups.

  4. 4.

To the best of our knowledge, for each of the property in the list of
consequences of Theorem D, there does not exist combination theorem which is
powerful enough to cover the pattern of combination of cyclic type and
spherical type Artin groups in Theorem F, thus all consequences are new for
the class of Artin groups in Theorem F. For example, the most recently
combination theorem for Farrell-Jones conjecture [Kno19] requires an
acylindrical action of the group on a tree, which is not satisfied in our
situation.

  5. 5.

Artin groups in Theorem F are in general not of type FC, so consequences of
Theorem D for this class does not follow from [HO21b].

These conditions are the most general that we can deal with. In particular, we
isolate simple families of Artin groups to which this result applies.

######  Corollary G.

Assume that $A_{\Gamma}$ is one of the following Artin groups:

  * $\bullet$

$A_{\Gamma}$ has rank at most $3$.

  * $\bullet$

$A_{\Gamma}$ is right-angled, without induced square.

Then $A_{\Gamma}\times\mathbb{Z}$ is Garside.

We emphasize that even for the simplest and extensively studied class of Artin
group, namely the class of right-angled Artin groups, not much is known about
the connection to Garside groups. Even for the free group $\mathbb{F}_{r}$ or
rank $r$, Bessis has defined a quasi-Garside structure on $\mathbb{F}_{r}$
([Bes06]). It is somehow striking that we are able to endow the direct product
$\mathbb{F}_{r}\times\mathbb{Z}$ with an actual Garside structure, and not a
mere quasi-Garside structure.

Assumptions of Theorem F have a close connection to an existing result for a
class of 2-dimensional Artin groups by [BM00]. More precisely, [BM00] studied
the class of _large type_ Artin groups, i.e. each edge in the Coxeter
presentation graph has label $\geq 3$. A dihedral subgroup of $A_{\Gamma}$ is
a subgroup generated by two vertices in an edge of $\Gamma$. Interestingly, if
we restrict Theorem F within the class of large type Artin groups, the left
forbidden configuration in [BM00, Figure 5] corresponds exactly to Assumption
1 in Theorem F, and the right forbidden configuration in [BM00, Figure 5]
corresponds exactly to Assumption 2 in Theorem F. There is a very interesting
geometric phenomenon behind this.

The strategy in [BM00] is to consider a dual Garside structure of each
dihedral subgroup (choosing a dual Garside structure amounts to choosing an
orientation of the associated edge), metrize each triangle in the presentation
complex with respect to the dual Garside structure as flat equilateral
triangles, and gluing these presentation complexes for dihedral subgroup in a
natural way to obtain a complex with fundamental group $A_{\Gamma}$. Then
[BM00, Thereom 7] implies that as long as the presentation graph $\Gamma$
avoids two configurations in [BM00, Figure 5], then resulting is locally
$\operatorname{CAT}(0)$.

Theorem F has a geometric counterpart (cf. Corollary 6.4). More precisely,
given an Artin group $A_{\Gamma}$, we can choose a dual Garside structure on
each standard spherical parabolic subgroups in a consistent way (again such
information can be encoded as an appropriate orientation of all large edges of
$\Gamma$). The dual Garside structure on each spherical parabolic subgroup $H$
gives an associated Garside complex (Definition 2.7) with fundamental group
$H$. By gluing these Garside complexes in a natural way, we obtain a complex
$X_{\Gamma}$ with fundamental group $A_{\Gamma}$. Here we metrize each simplex
in $X_{\Gamma}$ by a polyhedral norm which is related to the
$\widetilde{A}_{n}$-geometry, see [Hae22] (the norm here is not Euclidean),
which echoes the work of [BM00] where they metrize triangles with Euclidean
$\widetilde{A}_{2}$ shape. And the assumptions in Theorem F will ensure that
the universal cover of $X_{\Gamma}$ with such metric is a space with convex
geodesic bicombing (see Definition 2.12), which can be viewed as a form of
non-positive curvature, and echoes the CAT$(0)$ metric in [BM00].

It is natural to ask if we metrize each simplex in $X_{\Gamma}$ by Euclidean
simplices with $\widetilde{A}_{n}$-shape, whether the complex we obtain is
locally $\operatorname{CAT}(0)$. However, it is notoriously difficult to
verify local $\operatorname{CAT}(0)$-ness in high dimension. Here this issue
is bypassed through metrizing the simplices with different kinds of norm
rather then the Euclidean norm. While the resulting metric is not locally
$\operatorname{CAT}(0)$, it is almost as good as $\operatorname{CAT}(0)$ in
the sense that it implies most of the consequences of $\operatorname{CAT}(0)$
groups. We refer to [Hae22, Hae21], as well as [DL15, DL16] for more
discussion in this direction.

Interestingly, for every Artin group as in Theorem E, we have an answer to all
four questions stated above for general Artin groups. In particular, we can
deduce new cases of $K(\pi,1)$-conjecture from Theorem E.

######  Corollary H.

(=Corollary 5.8) Assume that $A_{\Gamma}$ is of hyperbolic cyclic type. Then
$A_{\Gamma}$ satisfies the $K(\pi,1)$ conjecture and has trivial center.

More precisely, the $K(\pi,1)$-conjecture is new for 6 examples of Artin
groups whose Coxeter groups act cocompactly on $\mathbb{H}^{3}$ or
$\mathbb{H}^{4}$. These examples seem to be rather difficult from the
viewpoint of other approaches of $K(\pi,1)$-conjecture. Though the
$K(\pi,1)$-conjecture when $A_{\Gamma}$ is 2-dimension hyperbolic cyclic type
follows from previous work [CD95], and there is also a more recent proof in
[DPS22] using dual quasi-Garside structures.

To put Corollary H in another context, note that the $K(\pi,1)$-conjecture is
proved by Artin groups associated with reflection groups acting on
$\mathbb{S}^{n}$ by Deligne [Del72], and Artin groups associated with
reflection groups acting on $\mathbb{E}^{n}$ by Paolini and Salvetti [PS21].
The next step is to look at Artin groups associated with reflection groups
acting on $\mathbb{H}^{n}$ (we call them hyperbolic type Artin groups), whose
$K(\pi,1)$-conjecture is widely open. A fundamental subclass of hyperbolic
type Artin groups are those associated with hyperbolic reflection groups whose
fundamental domain is a compact simplex. This subclass is classified by Lanner
[Lan50], which consists in infinitely many members in dimension 2 (whose
$K(\pi,1)$-conjecture is already understood [CD95, DPS22]), and in 14
remaining cases in higher dimension. From this perspective, Corollary H treats
6 out of these 14 remaining cases.

Corollary H also follows from another article of the second named author
[Hua23, Theorem 1.4], via an alternative approach to the
$K(\pi,1)$-conjecture. However, the method here establishes all the properties
in Theorem D for hyperbolic cyclic type Artin groups, which are not
consequences of [Hua23].

#### Structure of the article

In Section 2, we collect some background, notably on Garside groups, Artin
groups and nonpositively curved spaces. In Section 3, we discuss the general
criterion of making $G\times\mathbb{Z}$ a Garside group and prove Theorem A.
Then we discuss examples of $T(5)$ and systolic restricted presentation
groups. In Section 4, we adapt Theorem A to the special situation of Artin
groups, and produce a criterion of when an Artin group times $\mathbb{Z}$ is
Garside, see Proposition 4.2 and Corollary 4.4. In Section 5, we verify the
criterion in Proposition 4.2 and Corollary 4.4 for cyclic type Artin groups.
In Section 6 we treat more general Artin groups and prove Theorem F.

#### Acknowledgment

The authors would like to thank Anthony Genevois, Jon McCammond, Alex Martin,
Damian Osajda and Dani Wise for interesting discussions. The authors thank the
Centre de recherches mathématiques de Montréal for hospitality.

Thomas Haettel was partially supported by French project ANR-22-CE40-0004
GOFR. Jingyin Huang was partially supported by a Sloan fellowship.

##  2 Background

We start by giving references for Theorem D in the introduction, then we
collect background definitions and results concerning Artin groups, Garside
groups and nonpositively curved spaces.

###  2.1 Proof of Theorem D

We now give precise references for the various items of Theorem D from the
introduction, listing consequences for a group $G$ such that
$G\times\mathbb{Z}$ is Garside.

######  Proof.

  1. 1.

This is a consequence of [HO21b], see also [Hae21].

  2. 2.

This is a consequence of [Deh15, Proposition 3.25].

  3. 3.

This is a consequence of [Hae22, Corollary 9.8].

  4. 4.

This is a consequence of [HH22].

  5. 5.

This is a consequence of [Mos97]. For consequences of biautomaticity, see for
instance [BH99, Wen03, BD19].

  6. 6.

This is a consequence of [CCG+20, Proposition 7.10].

  7. 7.

This is a consequence of [LL07], see also [HO21a].

  8. 8.

Since $G$ acts geometrically on a metric space with a convex geodesic
bicombing, according to [KR17, Theorem 6.1], it satisfies the Farrell-Jones
conjecture with finite wreath products.

  9. 9.

Since $G$ acts geometrically on a metric space with a convex geodesic
bicombing, according to [FO20], it satisfies the coarse Baum-Connes
conjecture.

  10. 10.

This is a consequence of the Farrell-Jones conjecture and [BLR08, Theorem
0.12].

∎

###  2.2 Coxeter groups and Artin groups

We recall the definitions of Coxeter groups and Artin groups.

For every finite simple graph $\Gamma$ with vertex set $S$ and with edges
labeled by some integer in $\\{2,3,\dots\\}$, one associates the Coxeter group
$W(\Gamma)$ with the following presentation:

$W(\Gamma)=\langle S\,|\,\forall\\{s,t\\}\in\Gamma^{(1)},\forall s\in
S,s^{2}=1,[s,t]_{m}=[t,s]_{m}\mbox{ if the edge $\\{s,t\\}$ is labeled
$m$}\rangle,$

where $[s,t]_{m}$ denotes the word $ststs\dots$ of length $m$. Such a graph
$\Gamma$ may be called a _Coxeter presentation graph_ , emphasizing the fact
that edges correspond to relations.

We will also be using a graph closely related to $\Gamma$, the _Dynkin
diagram_ $\Gamma_{D}$: it has the same vertex set $S$, with some edges labeled
in $\\{4,5,\dots,\infty\\}$, with the following edges between vertices $s,t\in
S$:

  * $\bullet$

If there is an edge labeled $2$ between $s$ and $t$ in $\Gamma$, there is no
edge between $s$ and $t$ in $\Gamma_{D}$.

  * $\bullet$

If there is an edge labeled $3$ between $s$ and $t$ in $\Gamma$, there is an
unlabeled edge between $s$ and $t$ in $\Gamma_{D}$.

  * $\bullet$

If there is an edge labeled by $m\geqslant 4$ between $s$ and $t$ in $\Gamma$,
there is the same edge between $s$ and $t$ in $\Gamma_{D}$ labeled $m$.

  * $\bullet$

If there is no edge between $s$ and $t$ in $\Gamma$, there is an edge between
$s$ and $t$ in $\Gamma_{D}$ labeled $\infty$.

The associated Artin group $A(\Gamma)$ is defined by a similar presentation:

$A(\Gamma)=\langle
S\,|\,\forall\\{s,t\\}\in\Gamma^{(1)},[s,t]_{m}=[t,s]_{m}\mbox{ if the edge
$\\{s,t\\}$ is labeled $m$}\rangle.$

The groups $A(\Gamma)$ are also called Artin-Tits groups, since they have been
defined by Tits in [Tit66].

Note that only the relations $s^{2}=1$ have been removed, so that there is a
natural surjective morphism from $A(\Gamma)$ to $W(\Gamma)$. Also note that
when $m=2$, then $s$ and $t$ commute, and when $m=3$, then $s$ and $t$ satisfy
the classical braid relation $sts=tst$.

For a subset $S^{\prime}$ of the generating $S$, the subgroup of $A(\Gamma)$
or $W(\Gamma)$ generated by $S^{\prime}$ is called a _standard parabolic
subgroup_. A standard parabolic subgroup of an Artin group is itself an Artin
group [VdL83]. A similar statement is true for Coxeter groups [Bou02]. A
_parabolic subgroup_ is a conjugate of a standard parabolic subgroup.

Most results about Artin-Tits groups concern particular classes. The Artin
group $A(\Gamma)$ is called:

  * $\bullet$

of _spherical type_ if its associated Coxeter group $W(\Gamma)$ is finite,
i.e. may be realized as a reflection group of a sphere.

  * $\bullet$

of _Euclidean type_ if its associated Coxeter group $W(\Gamma)$ may be
realized as a reflection group of a Euclidean space.

  * $\bullet$

of _hyperbolic type_ if its associated Coxeter group $W(\Gamma)$ may be
realized as a reflection group of a real hyperbolic space.

We say a Coxeter group $W_{S}$ is of _cyclic type_ if the associated Dynkin
diagram is a cycle, and the parabolic subgroup generated by
$S\setminus\\{s\\}$ is spherical for any vertex $s\in\Gamma$. We list in Table
1 the Dynkin diagrams of cyclic type. Note that we use in this table the
convention of Dynkin diagrams: vertices that are not joined by an edge
commute, and we drop the label $3$ from edges. Note that cyclic type Coxeter
groups are either of Euclidean type or of hyperbolic type.

Name |  $\widetilde{A_{n}}$, for $n\geqslant 3$ | Triangle | $3-3-3-4$ | $3-3-3-5$
---|---|---|---|---
Dynkin diagram | $\dots$ | $\geqslant 3$$\geqslant 3$$\geqslant 3$ | $4$ | $5$
Name | $3-4-3-4$ | $3-4-3-5$ | $3-5-3-5$ | $3-3-3-3-4$
Dynkin diagram | $4$$4$ | $4$$5$ | $5$$5$ | $4$
Table 1: Diagrams of cyclic type

For an element $g$ in Coxeter group $W_{S}$, we can represent $g$ as a word in
the free monoid on $S$. Such representation is _reduced_ if its length is the
shortest possible among words in the free monoid that represent $g$. It is
known that any two reduced words representing the same element in $W_{S}$
differ by a finite sequence of moves applying the relation in $W_{S}$. Thus
each element in $W_{S}$ has a well-defined _support_ , which is the collection
of elements in $S$ which appears in a reduced word representing this element.

A subset $S^{\prime}\subset S$ is _irreducible_ if it spans a connected
subgraph of the Dynkin diagram, otherwise $S^{\prime}$ is _reducible_.

######  Lemma 2.1.

The support of each reflection is irreducible.

######  Proof.

Let $r=wsw^{-1}$ be a reflection in $W_{S}$ with $s\in S$ and $w\in W_{S}$. If
$\operatorname{Supp}(r)$ is reducible, then
$\operatorname{Supp}(r)=I_{1}\sqcup I_{2}$ with elements in $I_{1}$ commuting
with elements in $I_{2}$. As $\langle r\rangle$ is a parabolic subgroup of
$W_{S}$ which is contained in the standard parabolic subgroup $W_{I_{1}\cup
I_{2}}$, by [Qi07], there exists $w^{\prime}\in W_{I_{1}\cup I_{2}}$ and
$s^{\prime}\in I_{1}\cup I_{2}$ such that
$r=w^{\prime}s^{\prime}(w^{\prime})^{-1}$. We assume without loss of
generality that $s^{\prime}\in I_{1}$. Write
$w^{\prime}=w^{\prime}_{1}w^{\prime}_{2}$ with $w^{\prime}_{i}\in W_{I_{i}}$
for $i=1,2$. Then $r=w^{\prime}_{1}s^{\prime}(w^{\prime}_{1})^{-1}$ and
$\operatorname{Supp}(r)\subset I_{1}$, which is a contradiction. Thus the
lemma is proved. ∎

###  2.3 The $K(\pi,1)$-conjecture

Artin groups are closely related to hyperplane complements, which can be
presented in a simple way in spherical, Euclidean and hyperbolic types. Fix a
Coxeter group $W=W(\Gamma)$ of spherical type, Euclidean or hyperbolic type
acting by isometries on a sphere $\SS^{n-1}$, Euclidean space
$\mathbb{R}^{n-1}$ or a real hyperbolic space $\mathbb{H}^{n-1}$, where the
standard generators act by reflections.

In the case of $\SS^{n-1}$, we will consider $W$ as a subgroup of $O(n)$
acting by linear transformations on $\Omega=\mathbb{R}^{n}$. In the case of
$\mathbb{R}^{n-1}$, we will consider $W$ as a subgroup of
$\operatorname{GL}(n)$ acting by linear transformations on $\mathbb{R}^{n}$,
preserving the hyperplane $\\{x_{n}=1\\}$ and acting by isometries on it. The
group $W$ preserves the open cone $\Omega=\\{x_{n}>0\\}$ of $\mathbb{R}^{n}$.
In the case of $\mathbb{H}^{n-1}$, we will consider $W$ as a subgroup of
$O(n-1,1)$ acting linearly on $\mathbb{R}^{n}$, and preserving the open cone
$\Omega=\mathbb{H}^{n-1}$. A conjugate of an element of the standard
generating set $S$ is called a _reflection of $W$_. Let ${\cal R}$ denote the
set of reflections of $W$. Consider the family of linear hyperplanes of
$\mathbb{R}^{n}$

${\cal H}=\\{H_{r}\,|\,r\in{\cal R}\\},$

where $H_{r}\subset\mathbb{R}^{n}$ denotes the fixed point set of the
reflection $r$.

The analogue of the complement of the complexified hyperplane arrangement is

$M(\Gamma)=(\Omega\times\Omega)\char 92\relax\bigcup_{r\in{\cal
R}}(H_{r}\times H_{r}),$

see [Par14] for more details. Note that $W$ acts naturally on $M$, and we have
the following (see [VdL83]):

$\pi_{1}(W(\Gamma)\char 92\relax M(\Gamma))\simeq A(\Gamma).$

So the Artin group $A(\Gamma)$ appears as the fundamental group of (a quotient
of) the complement of a complexified hyperplane arrangement. One very natural
question is to decide whether it is a classifying space. This is the statement
of the following conjecture.

######  Conjecture ($K(\pi,1)$ conjecture).

The space $M(\Gamma)$ is aspherical.

This conjecture has been proved for spherical type Artin groups by Deligne in
[Del72], for 2-dimensional and type FC Artin groups by Charney and Davis in
[CD95], and for Euclidean type Artin groups by Paolini and Salvetti in [PS21]
very recently.

###  2.4 Interval groups and Garside groups

We will follow McCammond’s article [McC05] for the description of interval
groups.

######  Definition 2.2 (Posets).

A poset $P$ is called _bounded_ if it has a minimum, denoted $0$, and a
maximum, denoted $1$.

For $x\leq y$ in a poset $P$, the _interval_ between $x$ and $y$ is the
restriction of the poset to those elements $z$ with $x\leq z\leq y$. We denote
this interval by $[x,y]$. A poset $P$ is called _graded_ if for any $x\leq y$
in $P$, any chain in $[x,y]$ belongs to a maximal chain and all maximal chains
have the same finite length.

A poset $P$ is called _weakly graded_ if there is a poset map
$r:P\rightarrow\mathbb{Z}$, i.e. such that for every $x<y$ in $P$, we have
$r(x)<r(y)$: the map $r$ is called a _rank map_. A poset $P$ is called _weakly
boundedly graded_ if there is a rank map $r:P\rightarrow\mathbb{Z}$ with
finite image.

An _upper bound_ for a pair of elements $a,b\in P$ is an element $c\in P$ such
that $a\leq c,b\leq c$. A _minimal upper bound_ for $a,b$ is an upper bound
$c$ such that there does not exist upper bound $c^{\prime}$ of $a,b$ such that
$c^{\prime}<c$. The _meet_ of two elements $a,b$ in $P$ is an upper bound $c$
of them such that for any other upper bound $c^{\prime}$ of $a,b$, we have
$c\leq c^{\prime}$. We define _lower bound_ , _maximal lower bound_ , and
_join_ similarly. In general, the meet or join of two elements in $P$ might
not exist. A poset $P$ is a _lattice_ if any pair of elements have a meet and
a join.

A poset $P$ is a _meet-semilattice_ (resp. join-semilattice) if any pair of
elements have a meet (resp. a join).

######  Definition 2.3.

We say that a poset $P$ contains a _bowtie_ if there exist pairwise distinct
elements $a,b,c$ and $d$ such that $a,b<c,d$, and there exists no $x\in P$
such that $a,b\leqslant x\leqslant c,d$.

It turns out that bowties are the only obstruction to being a lattice, for a
weakly graded poset. This is proved in [BM10, Proposition 1.5] for bounded
graded lattices. This also holds for weakly graded lattices, so we give a
proof here for the convenience of the reader.

######  Proposition 2.4.

Let $L$ denote a weakly graded poset. Then $L\cup\\{0,1\\}$ is a lattice if
and only if $L$ has no bowtie.

######  Proof.

Assume that $L\cup\\{0,1\\}$ is a lattice, and consider $a,b<c,d$ in $L$. Then
the meet $x$ of $c,d$ is such that $a,b\leqslant x\leqslant c,d$. So $L$ has
no bowties.

Conversely, assume that $L$ has no bowtie. Note that $L\cup\\{0,1\\}$ has no
bowtie either. Fix $a,b\in L$, and let $M$ denote the set of upper bounds of
$a$ and $b$ in $L\cup\\{0,1\\}$: we have $1\in M$, so $M$ is not empty. Let us
consider a sequence $(x_{n})_{n\in\mathbb{N}}$ in $M$ such that for each
$n\in\mathbb{N}$, we have $x_{n}\geqslant x_{n+1}$. Let
$r:P\rightarrow\mathbb{Z}$ denote a weak grading on $P$. Then the sequence
$(r(x_{n}))_{n\in\mathbb{N}}$ in $\mathbb{Z}$ is non-increasing and bounded
below by $r(a)$, so it is eventually constant. This implies that the sequence
$(x_{n})_{n\in\mathbb{N}}$ itself is eventually constant.

We may therefore consider a minimal element $x$ of $M$. We will prove that $x$
is a unique: by contradiction, assume that $y\in M$ is a minimal element
distinct from $x$. Then $a,b<x,y$ form a bowtie. Hence $x$ is the unique
minimal element of $M$, and it is the join of $a$ and $b$ in $L\cup\\{0,1\\}$.

Similarly, any two elements of $L$ has a meet in $L\cup\\{0,1\\}$. So
$L\cup\\{0,1\\}$ is a lattice. ∎

Here is one definition of Garside groups. We refer the reader to [Deh15] and
[McC05] for more background on Garside groups. We also refer the reader to
[HH22] to equivalent definitions of Garside groups, which are more geometric
in flavour.

######  Definition 2.5 (Garside group).

Let $G$ denote a group, $S\subset G$ a finite subset and $\Delta\in G$. The
triple $(G,S,\Delta)$ is called a _Garside structure_ if the following
conditions hold. Let $G^{+}$ denote the submonoid of $G$ generated by $S$.

  1. 1.

The group $G$ is generated by $S$.

  2. 2.

For any element $g\in G^{+}$, there is a bound on the length of expressions
$g=s_{1}\dots s_{n}$, where $s_{1},\dots,s_{n}\in S\char 92\relax\\{1\\}$.

  3. 3.

We define the partial $\leqslant_{L}$, $\leqslant_{R}$ on $G^{+}$ by
$a\leqslant_{L}b$ if and only if $b=ac$ for some $c\in G^{+}$ and
$a\leqslant_{R}b$ if and only if $b=ca$ for some $c\in G^{+}$. The left
$\leqslant_{L}$ and right $\leqslant_{R}$ orders on $G^{+}$ are lattices.

  4. 4.

The set $S$ is a balanced interval between $1$ and $\Delta$, i.e.

$S=\\{g\in G^{+}\,|\,1\leqslant_{L}g\leqslant_{L}\Delta\\}=\\{g\in
G^{+}\,|\,1\leqslant_{R}g\leqslant_{R}\Delta\\}.$

A group is called _Garside_ if it admits such a Garside structure, and
$\Delta$ is called the _Garside element_. If the set $S$ is allowed to be
infinite, we may say that $(G,S,\Delta)$ is a _quasi-Garside structure_.

######  Definition 2.6 (Labeled posets).

If $P$ is a poset, the _set of intervals_ is $I(P)=\\{(x,y)\in
P^{2}\,|\,x\leqslant y\\}$.

Let $P$ denote a bounded poset, and let $S$ denote a labeling set.

An _interval-labeling_ of $P$ is a map $\lambda:I(P)\rightarrow S$.

An interval-labeling $\lambda$ is _group-like_ if, for any two chains
$x\leqslant y\leqslant z$ and $x^{\prime}\leqslant y^{\prime}\leqslant
z^{\prime}$ having two pairs of corresponding labels in common, the third pair
of labels are equal.

An interval-labeling $\lambda$ is _balanced_ if

$\\{\lambda(0,x)\,|\,x\in P\\}=\\{\lambda(x,1)\,|\,x\in
P\\}=\\{\lambda(x,y)\,|\,(x,y)\in I(P)\\}.$

Note that McCammond’s definition of balanced interval labeling ([McC05,
Definition 1.11]) only requires the first equality to hold. However, McCammond
states that the second inequality is a consequence of being balanced and
group-like, which does dot seem obvious. We therefore chose to strenghten the
definition of a balanced labeling, in order to ensure that all consequences of
a combinatorial Garside structure hold.

######  Definition 2.7 (Interval complex and interval group).

Let $P$ denote a poset with a group-like interval-labeling $\lambda$.

Let us consider the quotient $K_{P}$ of the geometric realization $|P|$ of
$P$, where the $k$-simplices corresponding to two $k$-chains
$(x_{0}<x_{1}<\dots<x_{k})$ and
$(x^{\prime}_{0}<x^{\prime}_{1}<\dots<x^{\prime}_{k})$ are identified if and
only if
$\lambda(x_{0},x_{1})=\lambda(x^{\prime}_{0},x^{\prime}_{1}),\dots,\lambda(x_{k-1},x_{k})=\lambda(x^{\prime}_{k-1},x^{\prime}_{k})$.
It is called the _interval complex_ of $P$.

The fundamental group $G_{P}$ of $K_{P}$ is called the _interval group_ of
$P$, it is naturally a quotient of the free group over $S$.

######  Example.

Let us consider the Boolean lattice $P={\cal P}(S)$ consisting of all subsets
of a finite set $S$. The geometric realization $|P|$ of $P$ is isomorphic to a
simplicial subdivision of the cube $[0,1]^{S}$.

For each $x\subset y\subset S$, let us consider the labeling
$\lambda(x,y)=y-x\in P$. The corresponding quotient $K_{P}$ is isomorphic to a
simplicial subdivision of the torus $(\SS^{1})^{S}$. The interval group
$G_{P}$ is isomorphic to the free abelian group $\mathbb{Z}^{S}$, with the
following presentation:

$G_{P}=\langle P\,|\,\forall x\subset y\subset z\subset
S,(y-x)\cdot(z-y)=(z-x)\rangle\simeq\mathbb{Z}^{S}.$

######  Definition 2.8 (Combinatorial Garside structure).

A _combinatorial Garside structure_ is a poset $P$ with an interval-labeling
$\lambda:I(P)\rightarrow S$ such that:

  * $\bullet$

$P$ is a (finite) bounded, weakly graded lattice.

  * $\bullet$

$\lambda$ is group-like and balanced.

If $P$ is infinite, we may say that it is a quasi-Garside.

Combinatorial Garside structures are just an explicit combinatorial way to
describe arbitrary Garside groups, as explained by McCammond.

######  Theorem 2.9.

[McC05, Theorem 1.17] A group $G$ is a Garside group if and only if $G$ is
isomorphic to the interval group of a finite combinatorial Garside structure.

######  Remark.

More generally, a group is quasi-Garside if and only if it is isomorphic to
the interval group of an arbitrary combinatorial Garside structure.

###  2.5 Nonpositive curvature: Helly graphs and CUB spaces

We will present briefly various notions of metric spaces and graphs of
nonpositive curvature which are relevant to Garside groups.

Let us start with Helly graphs: we refer the reader to [CCG+20] for more
details.

######  Definition 2.10 (Helly graph, Helly group).

A connected graph $\Gamma$ is called _Helly_ if any family of pairwise
intersecting combinatorial balls have a non-empty total intersection.

A group is called _Helly_ if it acts geometrically by automorphisms on a Helly
graph.

Helly groups enjoy many properties which are typical of nonpositive curvature,
see for instance [CCG+20], [Lan13] and [HO21a] and also Theorem D.

A much weaker, but way broader notion is that of weakly modular graphs, see
[CCHO21]. These graphs encompass many "nonpositive curvature type" graphs,
such as Helly graphs, (weakly) systolic graphs, median and quasi-median
graphs, modular graphs.

######  Definition 2.11 (Weakly modular graph).

A connected graph $\Gamma$ is called _weakly modular_ if it satisfies the
triangle condition (TC) and the quadrangle condition (QC):

  * (TC)

For any $x,y,z\in\Gamma^{(0)}$ such that $d(y,z)=1$ and
$d(x,y)=d(x,z)=n\geqslant 2$, there exists $t\in\Gamma^{(0)}$ such that
$d(t,y)=d(t,z)=1$ and $d(x,t)=n-1$.

  * (QC)

For any $x,y,z,u\in\Gamma^{(0)}$ such that $d(y,u)=d(z,u)=1$, $d(y,z)=2$,
$d(x,u)=n\geqslant 3$ and $d(x,y)=d(x,z)=n-1$, there exists $t\in\Gamma^{(0)}$
such that $d(t,y)=d(t,z)=1$ and $d(x,t)=n-2$.

Many of the consequences for Helly groups rely simply on the existence of a
convex geodesic bicombing, whose definition we recall here. We also recall the
definition of CUB spaces and groups, defined in [Hae22].

######  Definition 2.12 (Bicombing, CUB).

A _convex geodesic bicombing_ on a metric space $X$ is a map $\sigma:X\times
X\times[0,1]\rightarrow X$ such that:

  * $\bullet$

For each $x,y\in X$, the map $t\in[0,1]\mapsto\sigma(x,y,t)$ is a constant
speed reparametrized geodesic from $x$ to $y$.

  * $\bullet$

For each $x,x^{\prime},y,y^{\prime}\in X$, the map $t\in[0,1]\mapsto
d(\sigma(x,y,t),\sigma(x^{\prime},y^{\prime},t))$ is convex.

A metric space is called _CUB_ , for Convexly Uniquely Bicombable, if it
admits a unique convex geodesic bicombing. A group is called _CUB_ if it acts
geometrically by isometries on a CUB space.

Groups acting on spaces with convex bicombings enjoy many properties, see for
instance [DL16] and [DL15]. Furthermore, CUB groups satisfy some extra
properties presented in [Hae22], see also Theorem D.

One major incarnation of the nonpositive curvature properties of Garside
groups is the following.

######  Theorem 2.13 ([HO21b], see also [Hae21]).

Any Garside group acts geometrically by automorphisms on a Helly graph.

The quotient of a Garside group by the cyclic subgroup generated by the
Garside element also has nonpositive curvature in the following sense.

######  Theorem 2.14 ([Hae22],[HH22]).

Let $G$ denote a Garside group, with Garside element $\Delta$. Then the group
$G/\langle\Delta\rangle$ acts geometrically by isometries on a CUB space, and
it acts geometrically by automorphisms on a weakly modular graph.

###  2.6 Dual Garside structures on spherical type Artin groups

Dual Garside structure on spherical type Artin groups have been studied
notably by Birman-Ko-Lee ([BLR08]) and Bessis ([Bes03]), see also [Pao21] for
an overview of dual Garside structures on general Artin groups. We also refer
the reader to [McC05] for the point of view of interval groups that we are
presenting here.

Let $\Gamma$ denote a Coxeter presentation graph, with vertex set $S$. Given
any linear ordering $S=\\{s_{1},\dots,s_{n}\\}$ of $S$, we have an associated
_Coxeter element_ $\delta=s_{1}s_{2}\dots s_{n}$ in the Coxeter group
$W=W(\Gamma)$.

Let $R$ denote the set of _reflections_ of $W$, i.e. the set of all conjugates
of elements of $S$. Since $R$ generates $W$, we may consider its associated
word norm $\|\cdot\|_{R}$. In the Cayley graph of $W$ with respect to $R$, let
us consider the interval $P$ between $e$ and $\delta$: more precisely

$P=\\{u\in W\,|\,\|u\|_{R}+\|u^{-1}\delta\|_{R}=\|\delta\|_{R}=n\\}.$

The set $P$ has a natural partial (prefix) order $\leqslant_{L}$: if $u,v\in
P$, we say that $u\leqslant_{L}v$ if $\|u\|_{R}+\|u^{-1}v\|_{R}=\|v\|_{R}$.
Equivalently, $u$ is a prefix of a minimal expression of $v$ as a product of
reflections. Also equivalently, $u$ lies on a geodesic in the Cayley graph
between $e$ and $v$.

The poset $P$ is easily seen to be bounded and graded. Let us define an
interval-labeling $\lambda:I(P)\rightarrow W$ by $\lambda(u,v)=u^{-1}v\in W$:
this labeling is group-like and balanced. The poset $P$ is finite if and only
if $W$ is finite, i.e. if $\Gamma$ is of spherical type.

######  Definition 2.15 (Dual Artin group).

The _dual Artin group_ associated to $\Gamma$ and $\delta$ is the interval
group $A_{\delta}(\Gamma)$ of the poset $P$.

######  Theorem 2.16 (Birman-Ko-Lee [BKL98], Bessis [Bes03]).

If $\Gamma$ is of spherical type, for any Coxeter element $\delta$, the dual
Artin group $A_{\delta}(\Gamma)$ is isomorphic to the standard Artin group
$A(\Gamma)$. Moreover, the poset $P$ is a lattice: in particular, the Artin
group $A(\Gamma)$ is a Garside group.

###  2.7 Complexes associated with Garside groups

Consider a Garside group $G$, with positive monoid $G^{+}$, Garside element
$\Delta$ and Garside generating set $S$ as in Definition 2.5. Let $\leq_{L}$
and $\leq_{R}$ be the orders as in Definition 2.5, which also extend to orders
in $G$. More precisely, for $a,b\in G$, $a\leq_{L}b$ if $b=ac$ for some $c\in
G^{+}$, and $a\leq_{R}b$ if $b=ca$ for some $c\in G^{+}$.

The _Garside complex_ of $G$ is the simplicial complex $\widehat{X}_{G}$ with
vertex set $G$, and with simplices corresponding to chains
$g_{1}<_{L}g_{2}<_{L}\cdots<_{L}g_{n}$ such that $g_{n}\leq_{L}g_{1}\Delta$.
Note that $G$ acts properly and cocompactly by simplicial automorphisms on its
Garside complex. Alternatively, from the Garside group $G$, we can define an
associated combinatorial Garside structure with the underlying poset $P$ being
the set $\\{e\\}\cup S$ equipped with the the order $\leq_{L}$, and
$\lambda(x,y)=x^{-1}y$ for $x,y\in P$. Then the universal cover of the
interval complex associated with this combinatorial Garside structure is the
Garside complex.

The _Bestvina complex_ of $G$ is the simplicial complex $X_{G}$ whose vertices
are corresponds to cosets of $\langle\Delta\rangle$ in $G$ ([Bes99]). There is
an edge between two vertices in this complex if they have coset
representatives differ by right multiplication by element in
$S\setminus\\{\Delta\\}$, and the Bestvina complex is the flag complex induced
by this graph. Note that $\bar{G}=G/\langle\Delta\rangle$ acts properly and
cocompactly by simplicial automorphisms on the Bestvina complex. Topologically
$\widehat{X}_{G}$ is homeomorphic to $X_{G}\times\mathbb{R}$.

######  Theorem 2.17.

([Hae21, Theorem E]) For a Garside group $G$, if we metrize each simplex in
the Garside complex $\widehat{X}_{G}$ as orthoschemes with
$\ell^{\infty}$-metric, then $\widehat{X}_{G}$ is an injective metric space.
In particular it is CUB. Moreover, the injective metric on $\widehat{X}_{G}$
descends to a CUB metric on the Bestvina complex $X_{G}$, whose simplices are
equipped with special polyhedral norms in the sense of [Hae22].

##  3 Garside structure on $G\times\mathbb{Z}$

###  3.1 General construction

We will now present a general construction of a Garside structure on the
direct product $G\times\mathbb{Z}$, where $G$ is a group given by a specific
presentation with generating set denoted $U$. We will consider $U$ as an
abstract set endowed with a partial multiplication as defined below.

######  Definition 3.1 (Positive partial multiplication).

Let $U$ denote a set. A map $\cdot$ defined on a subset of $U\times U$ with
range $U$ is called a _positive partial multiplication_ if the following hold:

  * $\bullet$

Left associativity For any $u,v,w\in U$ such that $u\cdot v$ and $(u\cdot
v)\cdot w$ are defined, we require that $v\cdot w$ and $u\cdot(v\cdot w)$ are
defined, and that we have the equality $(u\cdot v)\cdot w=u\cdot(v\cdot w)$.

  * $\bullet$

Right associativity For any $u,v,w\in U$ such that $v\cdot w$ and
$u\cdot(v\cdot w)$ are defined, we require that $u\cdot v$ and $(u\cdot
v)\cdot w$ are defined, and that we have the equality $u\cdot(v\cdot
w)=(u\cdot v)\cdot w$.

  * $\bullet$

Identity There exists a distinguished element $e\in U$ such that, for every
$u\in U$, we have $e\cdot u=u\cdot e=u\in U$.

  * $\bullet$

Positivity For any $u,v\in U$ such that $u\cdot v=e$, we have $u=v=1$.

  * $\bullet$

Left cancellability For any $u,v,w\in U$ such that $u\cdot v=u\cdot w$, we
have $v=w$.

  * $\bullet$

Right cancellability For any $u,v,w\in U$ such that $v\cdot u=w\cdot u$, we
have $v=w$.

Let us define relations $\leqslant_{L},\leqslant_{R}$ on $U$ by:

$\displaystyle u\leqslant_{L}v$ if there exists $w\in U$ such that $u\cdot
w=v$ $\displaystyle u\leqslant_{R}v$ $\displaystyle\mbox{ if there exists
$w\in U$ such that $w\cdot u=v$}.$

######  Remark.

Given $u,v\in U$, we will often write in the sequel "$u\cdot v\in U$" in place
of "$u\cdot v$ is defined".

######  Lemma 3.2.

The relations $\leqslant_{L}$, $\leqslant_{R}$ are orders on $U$.

######  Proof.

By the existence of $1\in U$, we know that both relations are reflexive.

By the associativity assumption, we know that both relations are transitive.

We will now prove that $\leqslant_{L}$ is antisymmetric, the proof for
$\leqslant_{R}$ is similar. Let us assume that $u,v\in U$ are such that
$u\leqslant_{L}v$ and $v\leqslant_{L}u$. There exists $w,w^{\prime}\in U$ such
that $v=u\cdot w$ and $u=v\cdot w^{\prime}$, hence $u=(u\cdot w)\cdot
w^{\prime}=u\cdot(w\cdot w^{\prime})$ by associativity. Since $U$ is
cancellable, we deduce that $w\cdot w^{\prime}=e$. Since $U$ is positive, we
conclude that $w=w^{\prime}=e$, hence $u=v$. ∎

Note that the poset $(U,\leqslant_{L})$ admits an interval-labeling with
labels in $U$, i.e. for $u,v\in U$, the label of the interval between $u$ and
$u\cdot v$ is $v\in U$. One readily verifies that this interval-labeling is
group-like, so it makes sense to define the interval group $G_{U}$. In
particular, $G_{U}$ has the following presentation:

$G_{U}=\langle U\,|\,\forall u,v,w\in U\mbox{ such that }u\cdot v=w,\mbox{ we
have }uv=w\rangle.$

We will now describe the construction of a bounded poset $E$ consisting of two
"inverted" copies of $U$ as follows, which will be such that $G_{E}$ is
isomorphic to $G_{U}\times\mathbb{Z}$.

Let $\bar{U}$ be another copy of $U$, and we denote $\bar{u}\in\bar{U}$ to be
the element associated with $u\in U$. We will think $\bar{u}$ as a formal
inverse of $u$.

Consider the set $E=(U,0)\sqcup(\bar{U},1)$, with the following relation
$\preceq$:

  * $\bullet$

$(u,0)\preceq(v,0)$ if and only if $u\leqslant_{L}v$.

  * $\bullet$

$(u,0)\preceq(\bar{v},1)$ if and only if $v\cdot u\in U$.

  * $\bullet$

$(\bar{u},1)\preceq(\bar{v},1)$ if and only if $v\leqslant_{R}u$.

######  Lemma 3.3.

The relation $\prec$ is an order on $E$, with minimum $(e,0)$ and maximum
$(\bar{e},1)$.

######  Proof.

The reflexivity is clear. For transitivity, if $(u,0)\prec(\bar{v},1)$ and
$(\bar{v},1)\prec(\bar{w},1)$, then $v\dot{u}\in U$ and $w\leqslant_{R}v$.
Thus $v=w^{\prime}\cdot w$ for some $w^{\prime}\in U$. Thus $(w^{\prime}\cdot
w)\cdot u\in U$. By right associativity of the partial multiplication, we know
$w\cdot u\in U$. Thus $(u,0)\prec(\bar{w},1)$. Other cases of transitivity are
similar. The antisymmetry of $\prec$ follows from the antisymmetry of
$\leqslant_{L}$ and $\leqslant_{R}$ as in Lemma 3.2. ∎

Note that the poset $E$ is interval-labeled, with labels in $E$:

  * $\bullet$

For $u,v\in U$, the label of the interval between $(u,0)$ and $(u\cdot v,0)$
is $(v,0)\in E$.

  * $\bullet$

For $u,v,v\cdot u\in U$, the label of the interval between $(u,0)$ and
$(\bar{v},1)$ is $(\overline{v\cdot u},1)\in E$.

  * $\bullet$

For $u,v,v\cdot u\in U$, the label of the interval between $(\overline{v\cdot
u},1)$ and $(\bar{u},1)$ is $(v,0)\in E$.

######  Lemma 3.4.

The interval-labeled poset $E$ is group-like.

######  Proof.

Consider a chain with $3$ elements $a\prec b\prec c$ in $E$. Among the three
labels $\lambda(a,b)$, $\lambda(a,c)$ and $\lambda(b,c)$, we will show that
two of them determine the third one uniquely.

If $\lambda(a,b)$ and $\lambda(b,c)$ are known, there are three possibilities.

  * $\bullet$

Assume that $\lambda(a,b)=(u,0)$ and $\lambda(b,c)=(v,0)$. Then
$\lambda(a,c)=(u\cdot v,0)\in E$.

  * $\bullet$

Assume that $\lambda(a,b)=(u,0)$ and $\lambda(b,c)=(\bar{v},1)$. Then
$a=(w,0)$ for $w\in U$, $b=(w\cdot u,0)$ and $c=(\bar{x},1)$, where $x\cdot
w\cdot u\in U$. Then $\lambda(a,c)=(\overline{x\cdot w},1)\in E$.

  * $\bullet$

Assume that $\lambda(a,b)=(\bar{u},1)$ and $\lambda(b,c)=(v,0)$. Then
$a=(w,0)$ for some $w\in U$, $b=(\bar{x},1)$ and $c=(\bar{y},1)$ such that
$u=x\cdot w\in U$ and $y=v\cdot x\in U$. Since $a\prec c$, we know that
$y\cdot w\in U$, so $\lambda(a,c)=(\overline{y\cdot w},1)=(\overline{v\cdot
x\cdot w},1)=(\overline{v\cdot u},1)$.

If $\lambda(a,b)$ and $\lambda(a,c)$ are known, there are three possibilities.

  * $\bullet$

Assume that $\lambda(a,b)=(u,0)$ and $\lambda(a,c)=(v,0)$. Since $b\prec c$,
there exists $w\in U$ such that $u\cdot w=v$. Such $w$ is unique by
cancellability. Hence $\lambda(b,c)=(w,0)\in E$.

  * $\bullet$

Assume that $\lambda(a,b)=(u,0)$ and $\lambda(a,c)=(\bar{v},1)$. Then
$\lambda(b,c)=(\overline{v\cdot u},1)\in E$.

  * $\bullet$

Assume that $\lambda(a,b)=(\bar{u},1)$ and $\lambda(a,c)=(\bar{v},1)$. Then
$a=(w,0)$ for some $w\in U$, $b=(\bar{x},1)$ and $c=(\bar{y},1)$, with $x,y\in
U$ such that $x\cdot w=u$ and $y\cdot w=v$. Since $b\prec c$, there exists
$z\in U$ such that $z\cdot y=x$. Hence $z\cdot y\cdot w=x\cdot w$, so $z\cdot
v=u$. By cancellability, $z$ is uniquely determined by $u,v$. Then
$\lambda(b,c)=(\overline{z},1)$.

By symmetry, the remaining case is similar. ∎

######  Lemma 3.5.

The interval-labeled poset $E$ is balanced.

######  Proof.

The interval between $(u,0)$ and $(u\cdot v,0)$ has label $(v,0)\in E$, which
is also the label of the interval between $(e,0)$ and $(v,0)$, and also
between $(\bar{v},1)$ and $(\bar{e},1)$.

The interval between $(\overline{v\cdot u},1)$ and $(\bar{u},1)$ has label
$(\bar{v},1)\in E$, which is also the label of the interval between $(e,0)$
and $(\bar{v},1)$, and also between $(v,0)$ and $(\bar{e},1)$.

The interval between $(u,0)$ and $(\bar{v},1)$ has label $(v\cdot u,1)\in E$,
which is also the label of the interval between $(e,0)$ and $(\overline{v\cdot
u},1)$, and also between $(v\cdot u,0)$ and $(\bar{e},1)$. ∎

Given $u,v\in U$, a _left upper common bound_ for $u,v$ is a upper bound for
$\leq_{L}$. A _left join_ of $u$ and $v$ is an element $w\in U$ with
$u\leq_{L}w$ and $v\leq_{L}w$, such that $w\leq_{L}w^{\prime}$ for any other
left upper common bound $w^{\prime}$ of $u,v$. A left join, if exists, must be
unique. A _weak left join_ of $u$ and $v$ is an element $w\in U$ with
$u\leq_{L}w$ and $v\leq_{L}w$ such that there does not exist a left upper
common bound $w^{\prime}$ of $u,v$ such that $w^{\prime}<w$. Similarly, we
define right upper common bound and right (weak) join for $u,v$.

######  Proposition 3.6.

Let us consider the interval groups $G_{U},G_{E}$ associated with the
interval-labeled posets $U,E$. Then the natural map

$\displaystyle E$ $\displaystyle\mapsto$ $\displaystyle G_{U}\times\mathbb{Z}$
$\displaystyle(u,0)\in U\times\\{0\\}\subset E$ $\displaystyle\mapsto$
$\displaystyle(u,0)\in G_{U}\times\mathbb{Z}$ $\displaystyle(\bar{u},1)\in
U\times\\{0\\}\subset E$ $\displaystyle\mapsto$ $\displaystyle(u^{-1},1)\in
G_{U}\times\mathbb{Z}$

extends to an isomorphism of groups between $G_{E}$ and
$G_{U}\times\mathbb{Z}$.

######  Proof.

Note that $G_{U}\times\mathbb{Z}$ has generating set
$(U\times\\{0\\})\cup\\{(e,1)\\}$, and the relations are:

  1. 1.

$(e,1)(u,0)=(u,0)(e,1)$ for each $u\in U$;

  2. 2.

$(u,0)(v,0)=(w,0)$ for any $u,v,w\in U$ with $u\cdot v=w$.

On the other hand, the group $G_{E}$ has generating set $E$, and the relations
are:

  1. 1.

$(u,0)(v,0)=(w,0)$ for any $u,v,w\in U$ with $u\cdot v=w$;

  2. 2.

$(u,0)(\bar{v},1)=(\bar{w},1)$ for any $u,v,w\in U$ with $w\cdot u=v$;

  3. 3.

$(\bar{u},1)(v,0)=(\bar{w},1)$ for $u,v,w\in U$ with $v\cdot w=u$.

One readily checks that the map defined in the proposition extends to a group
homomorphism $G_{E}\to E_{U}$ as it is compatible with the relations.

We now define the inverse of this map on the standard generators of
$G_{U}\times\mathbb{Z}$:

$\displaystyle\left(U\times\\{0\\}\right)\cup\\{(e,1)\\}\subset
G_{U}\times\mathbb{Z}$ $\displaystyle\mapsto$ $\displaystyle G_{E}$
$\displaystyle(u,0)$ $\displaystyle\mapsto$ $\displaystyle(u,0)\in G_{E}$
$\displaystyle(e,1)$ $\displaystyle\mapsto$ $\displaystyle(\bar{e},1)\in
G_{E}.$

This map is also compactible with the relations of $G_{U}$ and $G_{E}$, note
that we see the second kind of relations of $G_{U}\times\mathbb{Z}$ in $G_{E}$
as follows:

$(u,0)(\bar{e},1)=(u,0)((\bar{u},1)(u,0))=((u,0)(\bar{u},1))(u,0)=(\bar{e},1)(u,0).$

Note that the composition of these two maps are identity on the generators,
thus they are inverses of each other. Then we are done. ∎

######  Proposition 3.7.

Assume that $U$ satisfies the following:

  1. 1.

$(U,\leqslant_{L})$ and $(U,\leqslant_{R})$ are weakly boundedly graded
posets.

  2. 2.

$(U,\leqslant_{L})$ and $(U,\leqslant_{R})$ are meet-semilattices.

  3. 3.

For any $a,u,v,w\in U$ such that $a\cdot u,a\cdot v\in U$ and $w$ is the join
for $\leqslant_{L}$ of $u$ and $v$, then $a\cdot w\in U$.

  4. 4.

For any $a,u,v,w\in U$ such that $u\cdot a,v\cdot a\in U$ and $w$ is the join
for $\leqslant_{R}$ of $u$ and $v$, then $w\cdot a\in U$.

  5. 5.

For any $a,b,u,v\in U$ such that $a\cdot u,a\cdot v,b\cdot u,b\cdot v\in U$,
either $a,b$ have a join for $\leqslant_{R}$, or $u,v$ have a join for
$\leqslant_{L}$.

Then $E$ is a lattice.

######  Proof.

Assumption 1 implies $E$ is a bounded graded poset. By Proposition 2.4, it is
sufficient to prove that $E$ contains no bowtie.

Assume that $(u,0),(v,0)\prec(w,0),(x,0)$ is a bowtie in $E$, where
$u,v,w,x\in U$: hence $u,v\leqslant_{L}w,x$ is a bowtie in $U$, which
contradicts that $(U,\leqslant_{L})$ is a meet-semilattice.

Assume that $(\bar{a},1),(\bar{b},1)\leqslant(\bar{c},1),(\bar{d},1)$ is a
bowtie in $E$, where $a,b,c,d\in U$: hence $a\in U\cdot c$, and
$c\leqslant_{R}a$. So $c,d\leqslant_{R}a,b$ is a bowtie in $U$, which
contradicts that $(U,\leqslant_{R})$ is a meet-semilattice.

Assume that $(u,0),(v,0)\prec(w,0),(\bar{a},1)$ is a bowtie in $E$, where
$u,v,w,a\in U$: hence $u,v\leqslant_{L}w$, so since we assumed to have a
bowtie, we have $w=u\vee_{L}v$. Also $a\cdot u,a\cdot v\in U$. By assumption,
this implies that $a\cdot w\in U$, so $(w,0)\prec(\bar{a},1)$.

Assume that $(a,0),(\bar{w},1)\prec(\bar{u},1),(\bar{v},1)$ is a bowtie in
$E$, where $u,v,w,a\in U$: hence $u,v\leqslant_{R}w$, so since we assumed to
have a bowtie, we have $w=u\vee_{R}v$. Also $u\cdot a,v\cdot a\in U$. By
assumption, this implies that $w\cdot a\in U$, so $(\bar{w},1)\prec(a,0)$.

Assume that $(u,0),(v,0)\prec(\bar{a},1),(\bar{b},1)$ is a bowtie in $E$,
where $u,v,a,b\in U$: hence $a\cdot u,a\cdot v,b\cdot u,b\cdot v\in U$. By
assumption, this implies that either $a,b$ have a join $c$ for $\leqslant_{R}$
or $u,v$ have a join $w$ for $\leqslant_{L}$. In each case, either
$(\bar{c},1)$ or $(w,0)$ is in the middle of the bowtie, which is a
contradiction. ∎

Let $K_{E}$ denote the interval complex of the labeled poset $E$.

######  Theorem 3.8.

Under the assumption of Proposition 3.7, the piecewise $\ell^{\infty}$ norm on
$K_{E}$ is injective and CUB. The group $G_{E}\simeq G_{U}\times\mathbb{Z}$ is
quasi-Garside with Garside element $(e,1)$ and set of simple elements $E$. If
$U$ is finite, then $E$ is finite and $G_{E}$ is Garside.

In particular, we can deduce the following result stated in the introduction.

######  Theorem 3.9.

Let $U$ be a finite set, endowed with a positive partial multiplication, and
let $G_{U}$ denote the associated interval group. Assume that the following
hold:

  * $\bullet$

$(U,\leqslant_{L})$ and $(U,\leqslant_{R})$ are semilattices.

  * $\bullet$

For any $a,u,v,w\in U$ such that $a\cdot u,a\cdot v\in U$ and $w$ is the join
for $\leqslant_{L}$ of $u$ and $v$, then $a\cdot w\in U$.

  * $\bullet$

For any $a,u,v,w\in U$ such that $u\cdot a,v\cdot a\in U$ and $w$ is the join
for $\leqslant_{R}$ of $u$ and $v$, then $w\cdot a\in U$.

  * $\bullet$

For any $a,b,u,v\in U$ such that $a\cdot u,a\cdot v,b\cdot u,b\cdot v\in U$,
either $a,b$ have a join for $\leqslant_{R}$, or $u,v$ have a join for
$\leqslant_{L}$.

Then the group $G_{U}\times\mathbb{Z}$ is a Garside group, with Garside
element $(e,1)$.

######  Remark.

We may remark that there are very simple situations where we can apply Theorem
3.9. For instance, let us consider the free group $F$ over a finite set $S$,
and let $U=S\cup\\{e\\}$. Then $U$ satisfies the assumptions of Proposition
3.7, and in particular the group $F\times\mathbb{Z}$ is Garside. This
particular case can also be deduced from [Pic22] via different methods. We
will however see, in the rest of the article, more interesting applications of
this result.

###  3.2 Some examples where Theorem 3.9 applies

######  Theorem 3.10.

Let us consider a group $G$ given by a finite presentation $\langle
S\,|\,r_{1}=r^{\prime}_{1},\dots,r_{n}=r^{\prime}_{n}\rangle$. Assume that the
following hold:

  * $\bullet$

For each $1\leqslant i\leqslant n$, the words $r_{i},r^{\prime}_{i}$ are
positive words in $S$, without common prefix or suffix.

  * $\bullet$

For each $1\leqslant i\leqslant n$, the word $r_{i}$ (and $r^{\prime}_{i}$)
does not appear as a subword of some of the other $2n-1$ words.

  * $\bullet$

For each distinct $s,t\in S$, there exist at most one $1\leqslant i\leqslant
n$ such the the first letters of $\\{r_{i},r^{\prime}_{i}\\}$ are $\\{s,t\\}$.

  * $\bullet$

For each distinct $s,t\in S$, there exist at most one $1\leqslant i\leqslant
n$ such the the last letters of $\\{r_{i},r^{\prime}_{i}\\}$ are $\\{s,t\\}$.

  * $\bullet$

The presentation is $T(5)$, i.e. the link of the vertex in the presentation
complex has girth at least $5$.

Then $G\times\mathbb{Z}$ is Garside.

######  Proof.

Let $U$ denote the quotient of the set of subwords of the words
$R=\\{r_{1},r^{\prime}_{1},\dots,r_{n},r^{\prime}_{n}\\}$ in $\mathbb{F}(S)$,
under the equivalence relation defined by $r_{i}\sim r^{\prime}_{i}$, for each
$1\leqslant i\leqslant n$. Given two positive words $u_{1},u_{2}$ of
$\mathbb{F}(S)$, we will write $u_{1}=u_{2}$ if they are the same word in
$\mathbb{F}(S)$, and $u_{1}\equiv u_{2}$ if they gives the same element in
$U$. Let us endow $U$ with the partial multiplication induced by the free
group $\mathbb{F}(S)$ on $S$. For any $u\in U$, let us define $u\cdot e\equiv
e\cdot u\equiv u$. For any $u,v\in U\char 92\relax\\{e\\}$, then $u\cdot v$
exists and is equal to $uv\in\mathbb{F}(S)$ if and only if $u,v\not\in R$ and
there exists $r\in R$ such that $uv$ is a subword of $r$.

We will show that this defines a positive partial multiplication on $U$.

Assume that $u,v,w\in U$ are such that $u\cdot v\in U$ and $(u\cdot v)\cdot
w\in U$. We will consider $u,v,w$ as representatives inside $\mathbb{F}(S)$.
Then there exists $r_{1}\in R$ such that $uv$ is a subword of $r_{1}$.
Moreover, $uv,w\notin R$ and $uvw$ is a subword of $r\in R$. Hence $vw$ is a
subword of $r$. Note that $v\notin R$, otherwise we will contradict the second
assumption of the theorem. Thus by the definition of product, $v\cdot w\equiv
vw\in U$. So $U$ is left associative, and similarly we can show it is right
associative.

The identity element $e\in\mathbb{F}(S)$ is an identity element for
$(U,\cdot)$. Now we check positivity. If $u\cdot v\in U$ and $u\cdot v\equiv
e$, then $u,v\notin R$ and $uv$ is a subword of $r\in R$. This forces $u=v=e$
as $u$ and $v$ are positive words. For the cancellability, if $u\cdot v\equiv
u\cdot w$, then either $uv=uw$ in $\mathbb{F}(S)$ and $v=w$ follows from the
cancellability in $\mathbb{F}(S)$, or there exists $i$ such that $uv=r_{i}$
and $uv=r^{\prime}_{i}$, which implies $u=e$ as we assume for each $1\leqslant
i\leqslant n$, the words $r_{i},r^{\prime}_{i}$ have no common prefix or
suffix.

So $U$ satisfies Definition 3.1. In particular, we may consider the left and
right orders on $U$.

Let us prove that $(U,\leqslant_{L})$ is weakly boundedly graded. It is clear
that $e\in U$ is the minimum of $U$. For each $u\in U$, let
$r(u)\in\mathbb{N}$ denote the maximal length of a representative for $u$ in
$\mathbb{F}(S)$. The map $r:U\rightarrow\mathbb{N}$ is a rank map with respect
to $\leqslant_{L}$ and $\leqslant_{R}$. Since $R$ is finite, $r$ has finite
image. So $(U,\leqslant_{L})$ and $(U,\leqslant_{R})$ are weakly boundedly
graded.

Let us prove that $(U,\leqslant_{L})$ is a semilattice. According to
Proposition 2.4, it is sufficient to prove that $(U,\leqslant_{L})$ does not
contain a bowtie. By contradiction, assume that $u,v\leqslant_{L}x,y$ is a
bowtie in $U$, with $r(x)-r(u)$ minimal, where $r:U\rightarrow\mathbb{N}$ is a
weak grading. Let $s,t\in S$ denote the first letters of $u,v$ respectively.
Since $r(x)-r(u)$ is minimal, we have $s\neq t$. We have
$s,t\leqslant_{L}x,y$. This implies that $x,y\in R$. By assumption, this
implies that $x=y$, so $u,v\leqslant_{L}x,y$ is not a bowtie: contradiction.
Hence $(U,\leqslant_{L})$ is a semilattice.

Let $a,u,v\in U$ such that $a\cdot u,a\cdot v\in U$ and $u\wedge_{L}v=w\in U$.
We want to prove that $a\cdot w\in U$. We will actually prove that
$u\leqslant_{L}v$ or $v\leqslant_{L}u$: if not, this means that there exists
$1\leqslant i\leqslant n$ such that $u,v$ are prefixes of
$r_{i},r^{\prime}_{i}$ respectively (up to switching $u$ and $v$). Then the
words $au,av,u^{-1}v$ give rise to a triangle in the link of the vertex in the
presentation complex, which contradicts the $T(5)$ assumption. So
$u\leqslant_{L}v$ or $v\leqslant_{L}u$, and hence $a\cdot w\in\\{a\cdot
u,a\cdot v\\}\subset U$.

Let us assume that $a,b,u,v\in U$ are pairwise distinct such that $a\cdot
u,a\cdot v,b\cdot u,b\cdot v\in U$. We will prove that either $a,b$ are
comparable for $\leqslant_{R}$, or $u,v$ are comparable for $\leqslant_{L}$.
If not, then the words $au,av,bv,bu$ give rise to a $4$-cycle in the link of
the vertex in the presentation complex, which contradicts the $T(5)$
assumption. So for instance $a,b$ are comparable for $\leqslant_{R}$, in which
case $a$ and $b$ have a join for $\leqslant_{R}$.

According to Proposition 3.7, we deduce that $G\times\mathbb{Z}$ is a Garside
group. ∎

######  Corollary 3.11.

For any surface $S$ of finite type (possibly non-orientable), except the
projective plane, $\pi_{1}(S)\times\mathbb{Z}$ is a Garside group.

######  Proof.

If $S$ is a surface with boundary, its fundamental group is a free group.

If $S$ is the torus, then $\pi_{1}(S)\simeq\mathbb{Z}^{2}$, which is a Garside
group, so $\pi_{1}(S)\times\mathbb{Z}\simeq\mathbb{Z}^{3}$ is a Garside group.

If $S$ is the closed orientable surface with genus $g\geqslant 2$, consider
the standard presentation

$G_{g}=\langle
a_{1},b_{1},\dots,a_{g},b_{g}\,|\,[a_{1},b_{1}]\dots[a_{g},b_{g}]=1\rangle.$

This presentation is not positive, so we will modify it as follows:

$\displaystyle G$ $\displaystyle=$ $\displaystyle\langle
a_{1},b_{1},h_{2},h_{3},\dots,h_{g-1},a_{g},b_{g}\,|\,a_{1}b_{1}h_{2}h_{3}\dots
h_{g-1}a_{g}b_{g}=a_{g}b_{g}h_{2}h_{3}\dots h_{g-1}a_{1}b_{1}$ $\displaystyle
a_{2}b_{2}=h_{2}b_{2}a_{2},\dots,a_{g-1}b_{g-1}=h_{g-1}b_{g-1}a_{g-1}\rangle.$

Then this presentation satisfies the assumptions of Theorem 3.10.

If $S$ is the projective plane, then $\pi_{1}(S)\simeq\mathbb{Z}/2\mathbb{Z}$,
so $\pi_{1}(S)\times\mathbb{Z}$ has torsion, hence it is not a Garside group.

If $S$ is the closed non-orientable surface with genus $2$, i.e. the Klein
bottle, then its fundamental group has the following presentation

$\pi_{1}(S)=\langle a,b\,|\,a^{2}=b^{2}\rangle,$

which is a Garside presentation with Garside element $\Delta=a^{2}=b^{2}$.
Hence $\pi_{1}(S)\times\mathbb{Z}$ is a Garside group.

If $S$ is the closed non-orientable surface with genus $g\geqslant 3$,
consider the (almost) standard presentation

$G=\langle a_{1},\dots,a_{g}\,|\,a_{1}^{2}\dots a_{g-1}^{2}=a_{g}^{2}\rangle,$

then it is easy to check that it satisfies the assumptions of Theorem 3.10. ∎

Another easy class of groups for which we can apply Theorem 3.8 is the
following class of groups with a systolic presentation. They have been defined
and studied by Soergel in [Soe21].

######  Definition 3.12 (Soergel [Soe21]).

A finite presentation $\langle S\,|\,R\rangle$ of a group $G$ is called a
_systolic restricted presentation_ if the following hold:

  * $\bullet$

Each relation $r\in R$ is of the form $r=abc^{-1}\in\mathbb{F}(S)$, where
$a,b,c\in S$.

  * $\bullet$

The flag completion of the Cayley graph of $G$ with respect to $S$ is
simplicial and systolic.

Note that asking that the Cayley graph of $G$ with respect to $S$ is
simplicial is equivalent to asking that any $s\in S$ has image in $G$
different from $e$, and also for any distinct $s,t\in S$, their image in $G$
are neither equal nor inverse. Soergel gives a complete characterization of
such systolic restricted presentations in [Soe21, Theorem 1].

Among Garside presentations, Soergel gives a characterization of those which
are systolic, see [Soe21, Theorem 2]. There are essentially amalgams of the
following Garside groups $G_{n,m}$, for $n,m\geqslant 1$, defined by the
following systolic restricted presentation:

$G_{n,m}=\langle x_{1},\dots,x_{n}\,|\,x_{1}x_{2}\dots x_{m}=x_{2}x_{3}\dots
x_{m+1}=\dots x_{n}x_{1}x_{2}\dots x_{m-1}\rangle.$

Among 2-dimensional Artin groups, Soergel gives a sufficient criterion in
terms of orientations of the edges of the Coxeter presentation graph, see
[Soe21, Theorem 3]. As a very restricted example, if the Coxeter presentation
graph $\Gamma$ has no triangles and no squares, then $A(\Gamma)$ admits a
systolic restricted presentation.

######  Theorem 3.13.

Let $G$ denote a group with a systolic restricted presentation. Then
$G\times\mathbb{Z}$ is a Garside group.

######  Proof.

Let us denote by $U=S\cup\\{e\\}$ in $G$, and let us consider the induced
partial multiplication from $G$. Since the Cayley graph of $G$ with respect to
$S$ is simplicial, we deduce that $U$ embeds in $G$.

The only non-trivial assumption to check for this partial multiplication is
the positivity: if there exist $s,t\in S$ such that $st=e$ in $G$, this
contradicts the fact that the Cayley graph of $G$ with respect to $S$ is
simplicial.

Since $U$ is finite, it is weakly boundedly graded, and it has minimum $e$.
Let us show that $(U,\leqslant_{L})$ is a meet-semilattice by contradiction:
let us assume that we have a bowtie $a,b<_{L}u,v$, with $a,b,u,v\in S$. Then
this corresponds to a loop of length $4$ in the link of the vertex $e$ in $X$.
By systolicity, we deduce that there exists a diagonal: either $a,b$ are
comparable, or $u,v$ are comparable. Hence $a,b<_{L}u,v$ is not a bowtie.
Similarly, $(U,\leqslant_{R})$ is a meet-semilattice

Let us now consider $a,u,v\in U$ such that $au,av\in U$ and $u,v$ have a join
$w\in U$ for $\leqslant_{L}$. Then $a^{-1},u,w,v$ form a loop of length $4$ in
the link of the vertex $e$ in $X$. By systolicity, we deduce that there exists
a diagonal: either $u,v$ are comparable, in which case $w\in\\{u,v\\}$ and
$aw\in\\{au,av\\}\subset U$, or there is an edge between $a^{-1}$ and $w$, in
which case $aw\in U$.

Let us now consider $a,b,u,v\in U$ such that $au,av,bu,bv\in U$. Then
$a^{-1},u,b^{-1},v$ form a loop of length $4$ in the link of the vertex $e$ in
$X$. By systolicity, we deduce that there exists a diagonal. If there is an
edge between $a^{-1}$ and $b^{-1}$, this means that $a$ and $b$ are
$\leqslant_{R}$-comparable, so they have a right join. If there is an edge
between $u$ and $v$, this means that $u$ and $v$ are
$\leqslant_{L}$-comparable, so they have a left join.

According to Theorem 3.8, we conclude that $G\times\mathbb{Z}$ is a Garside
group. ∎

######  Definition 3.14.

A finite presentation $\langle S\mid R\rangle$ is a _positive square
presentation_ if each relator $r$ is of form $ab=cd$ where $a,b,c,d$ are (not
necessarily distinct) elements in $R$.

Some natural examples of groups of square presentation include right-angled
Artin groups, mock right-angled Artin groups in the sense of [Sco08] and
groups arising from word labeled oriented graphs in the sense of [HR15]. We
give a criterion showing some of these groups are Garside groups after taking
the product with $\mathbb{Z}$.

Given a finite square presentation, let $X$ be the associated presentation
complex. Each edge loop of $X$ is oriented and labeled by an element in $S$.
Let $\Lambda$ be the link of the unique vertex of $X$. A vertex of $\Lambda$
is of type $o$ or $i$ if it corresponds to outgoing or incoming edge at the
base vertex.

######  Theorem 3.15.

Let $G$ denote a group with a positive square presentation such that the link
$\Lambda$ of its presentation complex satisfies the following conditions:

  1. 1.

there does not exist embedded 2-cycles in $\Lambda$ of type $(o,o)$ (means a
2-cycle with two vertices of type $o$) or type $(i,i)$;

  2. 2.

there does not exist embedded 3-cycles in $\Lambda$ of type $(o,o,i)$ or
$(i,i,o)$;

  3. 3.

there does not exist embedded 4-cycles in $\Lambda$ of type $(o,i,o,i)$.

Then $G\times\mathbb{Z}$ is a Garside group.

###### Proof.

Suppose the collection of relators are of form
$\\{a_{i}b_{i}=a^{\prime}_{i}b^{\prime}_{i}\\}_{i=1}^{k}$ where
$a_{i},b_{i},a^{\prime}_{i},b^{\prime}_{i}\in S$ for each $1\leq i\leq k$. Let
$U$ be the set equivalence classes of words in $\\{e\\}\cup
S\cup\\{a_{i}b_{i}\\}_{i=1}^{k}\cup\\{a^{\prime}_{i}b^{\prime}_{i}\\}_{i=1}^{k}$,
under the equivalence relation generated by $a_{i}b_{i}\sim
a^{\prime}_{i}b^{\prime}_{i}$ for $1\leq i\leq k$. We endow $U$ with the
partial multiplication as in the proof of Theorem 3.10. Now we verify the
assumptions of Theorem 3.9: $(U,\leq_{L})$ is a semilattice follows from the
lack of 2-cycle of type $(o,o)$ in $\Lambda$, as such kind of 2-cycles
correspond to bowties in $(U,\leq_{L})$; $(U,\leq_{R})$ is a semilattice
follows from the lack of 2-cycle of type $(i,i)$ in $\Lambda$. Now take
$a,u,v\in S$ with $u\neq v$ such that $au,av\in U$ and $u$ and $v$ have a left
join, then this gives a 3-cycle in $\Lambda$ made of vertices of type incoming
$a$, outgoing $u$, outgoing $v$, which is excluded by the lack of $3$-cycle of
type $(o,o,i)$. Similarly, the third item of Theorem 3.9 follows from the lack
of $3$-cycle of $(i,i,o)$. For the last item of Theorem 3.9, let $a,b,u,v\in
S$ with $u\neq v$ and $a\neq b$. If $au,av,bu,bv\in U$, this gives a 4-cycle
in $\Lambda$ with consecutive vertices of type $a$ incoming, $u$ outgoing, $b$
incoming and $v$ outgoing, which is ruled out by the lack of 4-cycle of type
$(o,i,o,i)$. Thus we are done by Theorem 3.9. ∎

##  4 Application to Artin groups

We will now explain how we can apply Theorem 3.9 for some Artin groups.

Let $(W,S)$ denote a Coxeter group, and let $A$ denote the associated Artin
group. Let $R$ denote the set of all reflections of $W$, and let $|\cdot|$
denote the reflection length on $W$. We will define the set $U$ inside $W$,
and we will look for conditions on $U$ and $W$ ensuring that the assumptions
from Theorem 3.9 are satisfied.

Given a subset $U\subset W$, we consider the following partial multiplication
$\cdot$ on $U$: if $u,v\in U$ are such that their product $uv$ in the Coxeter
group $W$ lies $U$ and furthermore $|u\cdot v|=|u|+|v|$, we define $u\cdot
v=uv\in U$. Let $R_{U}=R\cap U$.

######  Lemma 4.1.

Suppose $U$ satisfies the following conditions:

  1. 1.

For every $u\in U$, there exist $r_{1},\dots,r_{n}\in R_{U}$ such that
$u=r_{1}\cdot r_{2}\cdot\ldots\cdot r_{n}$.

  2. 2.

For every $r_{1},\ldots,r_{n}\in R_{U}$ such that $r_{1}\cdot
r_{2}\cdot\ldots\cdot r_{n}\in U$, we have $r_{1}\cdot r_{2}\cdot\ldots\cdot
r_{n-1}\in U$ and $r_{2}\cdot r_{3}\cdot\ldots\cdot r_{n}\in U$.

Then the set $(U,\cdot)$ satisfies Definition 3.1. In particular,
$\leqslant_{L}$ and $\leqslant_{R}$ are orders on $U$.

######  Proof.

It suffices to verify that for $u,v,w\in U$ such that $u\cdot v\cdot w\in U$,
we have $u\cdot v\in U$ and $v\cdot w\in U$. Indeed, by Assumption 1, let us
write reflection factorizations in $R_{U}$: $u=r_{1}\cdot
r_{2}\cdot\ldots\cdot r_{n}$, $v=r^{\prime}_{1}\cdot
r^{\prime}_{2}\cdot\ldots\cdot r^{\prime}_{n^{\prime}}$ and
$w=r^{\prime\prime}_{1}\cdot r^{\prime\prime}_{2}\cdot\ldots\cdot
r^{\prime\prime}_{n^{\prime\prime}}$. We then have $u\cdot v\cdot w\in U$, so
by assumption 2, we have both $u\cdot v\in U$ and $v\cdot w\in U$. ∎

We have a criterion for $E$ to be a lattice.

######  Proposition 4.2.

Assume that we have the following:

  1. 1.

For every $u\in U$, there exist $r_{1},\dots,r_{n}\in R_{U}$ such that
$u=r_{1}\cdot r_{2}\ldots r_{n}$.

  2. 2.

For every $r_{1},\ldots,r_{n}\in R_{U}$ such that $r_{1}\cdot
r_{2}\cdot\ldots\cdot r_{n}\in U$, we have $r_{1}\cdot r_{2}\cdot\ldots\cdot
r_{n-1}\in U$ and $r_{2}\cdot r_{3}\cdot\ldots\cdot r_{n}\in U$.

  3. 3.

For every $r_{1}\in R_{U}$ and $r_{2}\in R_{U}$ with a common left upper
bound, they have a left join; similarly, if $r_{1}$ and $r_{2}$ have a common
right upper bound, then they have a right join.

  4. 4.

For every $a\in U$, for any $u,v\in R_{U}$ such that $u,v$ have a left join
$w\in U$ and $a\cdot u,a\cdot v\in U$, we have $a\cdot w\in U$.

  5. 5.

For every $a\in U$, for any $u,v\in R_{U}$ such that $u,v$ have a right join
$w\in U$ and $u\cdot a,v\cdot a\in U$, we have $w\cdot a\in U$.

  6. 6.

For every $a,b,u,v\in R_{U}$ and any $x\in U$ such that $a\cdot x\cdot
u,a\cdot x\cdot v,b\cdot x\cdot u,b\cdot x\cdot v\in U$, we have that either
$u,v$ have a left join, or $a,b$ have a right join.

Then $U$ satisfies all the conditions in Proposition 3.7. In particular $E$ is
a lattice.

######  Proof.

By Lemma 3.2, it remains to verify that $U$ satisfies the assumptions of
Proposition 3.7.

Assumption 1 of Proposition 3.7 follows by considering the reflection length
on $U$.

We will now prove that $(U,\leqslant_{L})$ is a meet-semilattice. We
artificially add an largest element $\hat{1}$ to $U$, so
$P=(U\cup\\{\hat{1}\\},\leq_{L})$ is a bounded poset of finite length (i.e.
there is a finite upper bound on the lengths of its chains). Recall that an
element $p_{1}\in P$ _covers_ $p_{2}\in P$ if $p_{1}>p_{2}$ and there does not
exist $p\in P$ with $p_{1}>p>p_{2}$. We claim that if $u_{1},u_{2}$ are two
distinct elements in $P$ that covers $v$, then $u_{1}$ and $u_{2}$ has a join.
By assumption 2, we can write $u_{1}=v\cdot r_{1}$ and $u_{2}=v\cdot r_{2}$
with $r_{1},r_{2}\in R_{U}$. If $\hat{1}$ is the only common left upper bound
of $u_{1}$ and $u_{2}$, then clearly they have a join. If $u_{1}$ and $u_{2}$
have a common left upper bound $u^{\prime}$ other than $\hat{1}$, then we can
write $u^{\prime}=u_{i}\cdot w_{i}$ with $w_{i}\in U$ for $i=1,2$. Thus
$r_{1}$ and $r_{2}$ has a common left upper bound, which is
$w^{\prime}=r_{1}\cdot w_{1}=r_{2}\cdot w_{2}\in U$. Let $r$ be the left join
of $r_{1}$ and $r_{2}$. Then $r\leq_{L}w^{\prime}$, which implies that $v\cdot
r\leq_{L}v\cdot w^{\prime}=u^{\prime}$. Thus $v\cdot r$ is the join for
$u_{1}$ and $u_{2}$. Now it follows from [BEZ90, Lemma 2.1] that $P$ is a
lattice. Thus $(U,\leq_{L})$ is a meet-semilattice. Similarly we can prove
$(U,\leq_{R})$ is a meet-semilattice.

We now prove Assumption 3 of Proposition 3.7, i.e. for all $a,u,v\in U$ such
that $u,v$ have a join $w\in U$ for $\leqslant_{L}$, and $a\cdot u,a\cdot v\in
U$, then $a\cdot w\in U$.

We will prove it by decreasing induction on $|a|$, and for a fixed value of
$|a|$ by increasing induction on $|u|+|v|$. Since $U$ is finite, if $a$ is
maximal, then $u,v=e$, so the property is true. Now consider $a,u,v\in U$, and
assume that the property holds for any larger value of $|a|$. If $u,v\in
R_{U}$, the property holds by assumption. So let us assume that the property
holds for smaller values of $|u|+|v|$. Let us assume that $u\not\in R_{U}$,
and write $u=u_{1}\cdot r$, with $u_{1}\in U$, $r\in R_{U}$ and
$|u_{1}|=|u|-1$. According to Properties 1 and 2, we know that $a\cdot
u_{1}\in U$. Since $u_{1},v$ have an upper bound $w$, and since $U$ is a meet-
semilattice, they have a left join $w_{1}=u_{1}\cdot w^{\prime}$, with
$w^{\prime}\in U$. Since $|u_{1}|+|v|<|u|+|v|$, we deduce by induction that
$a\cdot w_{1}=a\cdot u_{1}\cdot w^{\prime}\in U$. Now $u_{1}\cdot w^{\prime}$
and $u=u_{1}\cdot r$ have an upper bound $w\in U$, so we deduce by Properties
1 and 2 that $w^{\prime}$ and $r$ have an upper bound in $U$, hence they also
have a join: let us write $w^{\prime}\vee_{L}r=w^{\prime\prime}\in U$. Since
$|au_{1}|>|a|$, we deduce by induction that $au_{1}\cdot w^{\prime\prime}\in
U$, in particular
$|au_{1}w^{\prime\prime}|=|au_{1}|+|w^{\prime\prime}|=|a|+|u_{1}|+|w^{\prime\prime}|$.
Note that $u=u_{1}r\leq_{L}u_{1}\cdot w^{\prime\prime}$ and
$v\leq_{L}w_{1}=u_{1}w^{\prime}\leq_{L}u_{1}w^{\prime\prime}$, we know
$w\leqslant_{L}u_{1}\cdot w^{\prime\prime}$, hence $aw\in U$ by Property 2. On
the other hand, $u^{-1}_{1}w$ is a left common upper bound for $r$ and
$w^{\prime}$. Hence $w^{\prime\prime}\leq_{L}u^{-1}_{1}w$ and $u_{1}\cdots
w^{\prime\prime}\leq_{L}w$. Then $w=u_{1}\cdots w^{\prime\prime}$. In
particular
$|aw|=|au_{1}w^{\prime\prime}|=|a|+|u_{1}|+|w^{\prime\prime}|=|a|+|w|$. Thus
$a\cdots w\in U$.

Assumption 4 of Proposition 3.7 can be proved in a similar way.

We will now prove Assumption 5 of Proposition 3.7, i.e. for every
$a,b,u,v,x\in U$ such that $a\cdot x\cdot u,a\cdot x\cdot v,b\cdot x\cdot
u,b\cdot x\cdot v\in U$, we have that either $u,v$ have a left join, or $a,b$
have a right join.

We will prove it by decreasing induction on $|x|$, and for a fixed value of
$|x|$ by increasing induction on $|a|+|b|+|u|+|v|$. Since $U$ is finite, if
$x$ is maximal, then $a,b,u,v$ are all equal to $e$, so the property is true.
Now consider $a,b,u,v,x\in U$, and assume that the property holds for any
larger value of $|x|$. If $a,b,u,v\in R_{U}$, then the property holds by
assumption. Without loss of generality, assume that $a\in U\char 92\relax
R_{U}$, and write $a=r\cdot a^{\prime}$, for some $r\in R_{U}$ and
$a^{\prime}\in U\char 92\relax\\{e\\}$ so that $|a^{\prime}|=|a|-1$. So
$a^{\prime}\cdot x\cdot u,a^{\prime}\cdot x\cdot v,b\cdot x\cdot u,b\cdot
x\cdot v\in U$: since $|a^{\prime}|<|a|$, we deduce by induction that either
$a^{\prime},b$ have a right join or $u,v$ have a left join, and in the latter
case we have the desired conclusion. Let us then assume that $a^{\prime},b$
have a right join $c\in U$. Let us write $c=c^{\prime}\cdot a^{\prime}$, where
$c^{\prime}\in U$.

Since $a^{\prime}\cdot xu,b\cdot xu\in U$ and $a^{\prime},b$ have a right join
$c\in U$, according to Property 5, we deduce that $c\cdot xu\in U$, and
similarly $c\cdot xv\in U$. We now consider the four elements $c^{\prime}\cdot
a^{\prime}x\cdot u=cxu,c^{\prime}\cdot a^{\prime}x\cdot v=cxv,r\cdot
a^{\prime}x\cdot u,r\cdot a^{\prime}x\cdot v$ in $U$. Since $a^{\prime}\cdot
x\in U$ and $|a^{\prime}x|>|x|$, we deduce by induction that either
$c^{\prime},r$ have a right join or $u,v$ have a left join, and in the latter
case we have the desired conclusion. Let us then assume that $c^{\prime},r$
have a right join $d\in U$. Since $c^{\prime}\cdot a^{\prime}=c,r\cdot
a^{\prime}=a\in U$ and $c^{\prime},r$ have a right join $d\in U$, according to
Property 5, we deduce that $d\cdot a^{\prime}\in U$. Now remark that
$a=ra^{\prime}\leqslant_{R}da^{\prime}$ and
$b\leqslant_{R}c=c^{\prime}a^{\prime}\leqslant_{R}da^{\prime}$, so $a,b$ have
a common right upper bound for $\leqslant_{R}$. Since $(U,\leqslant_{R})$ is a
semilattice, we conclude that $a,b$ have a right join. ∎

Let us denote by $K_{E}$ the interval complex of the poset $E$ as in
Definition 2.7, and let $G_{E}$ denote the corresponding interval group. We
will find a simple criterion ensuring that the interval group $G_{E}$ is
isomorphic to $A\times\mathbb{Z}$, where $A$ is the Artin group associated to
$W$.

######  Theorem 4.3.

Assume that, for each spherical $T\subset S$, there is a choice of Coxeter
element $w_{T}\in W_{T}$ such that, for every spherical $T^{\prime}\subset T$,
we have $w_{T^{\prime}}\leqslant_{L}w_{T}$. Assume that

$U=\bigcup_{T\subset S\mbox{ spherical}}[e,w_{T}].$

Then $K_{E}$ has the homotopy type of the Salvetti complex of the Artin group
$A\times\mathbb{Z}$. In particular, the interval group $G_{E}$ is isomorphic
to $A\times\mathbb{Z}$.

######  Proof.

For each spherical $T\subset S$, let us denote $U_{T}=[e,w_{T}]\subset U$.
Consider the subposet
$E_{T}=\left(U_{T}\times\\{0\\}\right)\sqcup\left(\overline{U_{T}}\times\\{1\\}\right)\subset
U$, and denote by $K_{E_{T}}\subset K_{E}$ the subcomplex corresponding to the
quotient of the geometric realization of $E_{T}$.

We claim that $K_{E_{T}}$ has the homotopy type of the Salvetti complex
$X_{T}$ of the Artin group $A_{T}\times\mathbb{Z}$. Indeed, let us denote by
$s_{0}\in A_{T}\times\mathbb{Z}$ a generator of $\mathbb{Z}$, so that the
Artin group $A_{T}\times\mathbb{Z}$ has standard generating system
$T^{\prime}=T\cup\\{s_{0}\\}$. Now $w^{\prime}_{T}=w_{T}s_{0}$ is a Coxeter
element for the spherical Coxeter group
$W^{\prime}_{T}=W_{T}\times\mathbb{Z}/2\mathbb{Z}$, and $K_{E_{T}}$ coincides
with the dual Salvetti complex for $w^{\prime}_{T}$ as described in [PS21,
Section 5]. According to [PS21, Remark 5.4], we deduce that $K_{E_{T}}$ has
the same homotopy type as the standard Salvetti complex $X_{T}$ for the
spherical Artin group $A_{T}\times\mathbb{Z}$.

By assumption on $U$, it is clear that $K$ is equal to the union of all
$K_{E_{T}}$, for $T\subset S$ spherical. Also remark that the standard
Salvetti complex $X$ for the Artin group $A\times\mathbb{Z}$ is equal to the
union of all $K_{E_{T}}$, for $T\subset S$ spherical.

According to the proof of [PS21, Theorem 5.5], we deduce that $K_{E}$ has the
homotopy type of $X$.

In particular, the interval group $G_{E}$ of $E$, which is the fundamental
group of $K_{E}$, is naturally isomorphic to the Artin group
$A\times\mathbb{Z}$. Moreover, the standard Salvetti complex $X$ of
$A\times\mathbb{Z}$ is aspherical, so in particular the standard Salvetti
complex of $A$ itself is aspherical: we deduce that the $K(\pi,1)$ conjecture
holds for $A$. ∎

######  Corollary 4.4.

Assume that $W$ is a Coxeter group, with a subset $U\subset W$ satisfying the
conditions of Proposition 4.2 and of Theorem 4.3. Let $A$ be the Artin group
associated with $W$. Then $A\times\mathbb{Z}$ is Garside, with Garside element
$(e,1)$. Moreover, the $K(\pi,1)$ conjecture holds for $A$.

######  Remark.

There are some Artin groups for which it is not possible to find a subset
$U\subset A$ satisfying the conditions of Proposition 4.2 and of Theorem 4.3.
Here are two simple examples.

  1. 1.

Consider the right-angled Artin group
$A\simeq\mathbb{F}_{2}\times\mathbb{F}_{2}$ with defining graph a square with
vertices $a,u,b,v$ in this cyclic order (see Figure 2), and assume that the
conditions of Theorem 4.3 hold. Then we have $au,av,bu,bv\in U$, but neither
$a,b$ nor $u,v$ have a join for $\leqslant_{L}$. Then the conditions of
Proposition 4.2 do not hold.

$a$$u$$b$$v$ Figure 2: The right-angled Artin group $A$ over a square.

  2. 2.

Consider the $A$ with defining graph a complete graph over $7$ vertices, whose
Dynkin diagram is a line with vertices $s_{1},s_{2},\dots,s_{7}$, with all
edge labels equal to $4$ (see Figure 3). Assume that the conditions of Theorem
4.3 hold. Consider the four elements of $U$: $a=s_{1}s_{2}$ or $s_{2}s_{1}$
(depending on the ordering on $S$), $b=s_{2}s_{3}$ or $s_{3}s_{2}$,
$u=s_{5}s_{6}$ or $s_{6}s_{5}$ and $v=s_{6}s_{7}$ or $s_{7}s_{6}$. Then
$au,av,bu,bv\in U$, but neither $a,b$ nor $u,v$ have a join for
$\leqslant_{L}$. Then the conditions of Proposition 4.2 do not hold.

$4$$4$$4$$4$$4$$4$$s_{1}$$s_{2}$$s_{3}$$s_{4}$$s_{5}$$s_{6}$$s_{7}$$a$$b$$u$$v$
Figure 3: The Dynkin diagram of an Artin group for which Corollary 4.4 does
not apply.

######  Corollary 4.5.

Assume that, for each spherical $T\subset S$, there is a choice of Coxeter
element $w_{T}\in W_{T}$ such that, for every spherical $T^{\prime}\subset T$,
we have $w_{T^{\prime}}\leqslant_{L}w_{T}$. Assume that

$U=\bigcup_{T\subset S\mbox{ spherical}}[e,w_{T}].$

Let $\widehat{U}$ be the lift of $U$ from $W_{S}$ to $A_{S}$ via the
(compatible) isomorphism between the dual Artin group associated with $A_{T}$
for each $T\subset S$ spherical and $A_{T}$ (cf. Theorem 2.16, and more
precisely [Bes03, Theorem 2.2.5]).

Assume that $U$ satisfying the conditions of Proposition 4.2. Let $X_{S}$ be
the flag complex of the Cayley graph of $A_{S}$ with generating set
$\widehat{U}$. Then $X_{S}$ admits an $A_{S}$-equivariant CUB metric such that
each simplex of $X_{S}$ is equipped with a polyhedral norm as in [Hae22].

######  Proof.

By Corollary 4.4, $A_{S}\times\mathbb{Z}$ is a Garside group with the choice
of fundamental interval $E$ as in Section 3. Note that Bestvina complex (cf.
Section 2.7) for the Garside group $A_{S}\times\mathbb{Z}$ is isomorphic to
flag complex of the Cayley graph of $A_{S}$ with generating set $\widehat{U}$.
Thus we are done by Theorem 2.17. ∎

##  5 Cyclic-type Artin groups

We will now describe a family of Artin groups for which we can find a set $U$
satisfying the conditions of Proposition 4.2 and of Theorem 4.3.

###  5.1 Spherical Artin group with linear Dynkin diagram

######  Lemma 5.1.

Let $W_{S}$ be an arbitrary Coxeter group (not necessarily spherical). Let $w$
denote a word in $S$ representing the trivial element of $W_{S}$. Then each
letter of $w$ appears at least twice.

######  Proof.

By contradiction, assume that we can write $w=usv$, where $s\in S$ and $u,v$
are words in $S\char 92\relax\\{s\\}$. Then, in the Coxeter groups $W_{S}$,
the words $s$ and $u^{-1}v^{-1}$ represent the same element. Since $s$ in the
support of $s$ and not of $u^{-1}v^{-1}$, this is a contradiction. ∎

######  Lemma 5.2.

Let $W_{S}$ be an arbitrary Coxeter group (not necessarily spherical). Let
$s\in S$, and let $w$ denote a reduced word in $S\char 92\relax\\{s\\}$
representing an element commuting with $s$. Then every letter of $w$ commutes
with $s$.

######  Proof.

Since $w$ and $s$ have disjoint supports, the words $sw$ and $ws$ are reduced.
We can pass from the reduced word $sw$ to the reduced word $ws$ by applying
only standard relations (see for instance [Dav15, Theorem 3.4.2]). This
implies that $s$ commutes with every letter of $w$. ∎

######  Lemma 5.3.

Let $W_{S}$ be an arbitrary Coxeter group (not necessarily spherical). Let
$\\{s_{1},s_{2},\ldots,s_{n}\\}\subset S$ such that, for each $1\leqslant
i\leqslant n-1$, there exists $i<j\leqslant n$ such that $s_{i}$ and $s_{j}$
do not commute. Then the word

$s_{1}\cdots s_{n-1}s_{n}s_{n-1}\cdots s_{1}$

is reduced.

######  Proof.

We induct on $n$. Then case $n=1$ is trivial. For the general case, by
contradiction, assume that the word $w=s_{1}\cdots s_{n-1}s_{n}s_{n-1}\cdots
s_{1}$ is not reduced. According to the deletion condition (see for instance
[Hum75, Theorem 5.8]), $w$ can also be represented by a word $w^{\prime}$
obtained from $w$ by deleting two letters.

Since $w$ represents a reflection of $W_{S}$, $w^{\prime}$ also represents a
reflection. According to the strong exchange condition (see [Hum75, Theorem
5.8]), if we remove one letter from $w^{\prime}$ we may obtain the trivial
element. According to Lemma 5.1, we deduce that there exists $1\leqslant
i\leqslant n-1$ such that $w^{\prime}$ is obtained from $w$ by removing the
two occurences of $s_{i}$.

So we have $w^{\prime}=s_{1}s_{2}\dots s_{i-1}s_{i+1}\dots s_{n}\dots
s_{i+1}s_{i-1}\dots s_{1}$. By conjugating by $s_{1}s_{2}\dots s_{i-1}$, we
deduce that the words $s_{i}\cdots s_{n-1}s_{n}s_{n-1}\cdots s_{i}$ and
$s_{i+1}\cdots s_{n-1}s_{n}s_{n-1}\cdots s_{i+1}$ represent the same element
of $W_{S}$. In particular, the element $u=s_{i+1}\cdots
s_{n-1}s_{n}s_{n-1}\cdots s_{i+1}$ commutes with $s_{i}$. As $u=s_{i+1}\cdots
s_{n-1}s_{n}s_{n-1}\cdots s_{i+1}$ is reduced by induction assumption,
according to Lemma 5.2, we deduce that $s_{i}$ commutes with every letter
$s_{i+1},\dots,s_{n}$, which contradicts the assumption. ∎

######  Lemma 5.4.

Suppose $S$ is spherical with linear Dynkin diagram. We label elements of $S$
as $\\{s_{1},\ldots,s_{n}\\}$ using a linear order of coming from the Dynkin
diagram. Let $U$ be the dual Garside interval with respect to the dual Garside
element $\delta=s_{1}s_{2}\cdots s_{n}$ and let $R_{U}$ be the set of
reflections in $U$. Assume that $u,v\in R_{U}$ are such that $u\cdot v\in U$.
Let $I=\operatorname{Supp}(u)$ and $J=\operatorname{Supp}(v)$. If
$\min(I)-1\in J$, then $I\subset J$.

######  Proof.

Up to symmetries, the Coxeter group $W_{S}$ is one of the following:

  * $\bullet$

Type $I_{m}$ with $m\geq 3$, and $n=2$.

  * $\bullet$

Type $A_{n}$.

  * $\bullet$

Type $B_{3}$, with Dynkin diagram $3-4$, and $n=3$.

  * $\bullet$

Type $H_{3}$, with Dynkin diagram $3-5$, and $n=3$.

  * $\bullet$

Type $B_{4}$, with Dynkin diagram $3-3-4$, and $n=4$.

  * $\bullet$

Type $F_{4}$, with Dynkin diagram $3-4-3$, and $n=4$.

Let us denote $I=\operatorname{Supp}(u)$ and $J=\operatorname{Supp}(v)$. Up to
passing to a standard parabolic subgroup, we can assume $S=I\cup J$.

Assume first that $|S|=2$, i.e. $W_{S}$ is of type $I_{m}$ with $m\geq 3$.
Assume that $I,J\neq S=\\{s_{1},s_{2}\\}$, so that $u,v\in\\{s_{1},s_{2}\\}$.
We only have to prove that $s_{2}s_{1}\not\leqslant_{L}\delta$.

By contradiction, assume that $s_{2}s_{1}\leqslant_{L}\delta$. Since
$\delta=s_{1}s_{2}$ has reflection length $2$, we deduce that
$s_{2}s_{1}=\delta$, so $s_{1}s_{2}=s_{2}s_{1}$. This contradicts $m\geqslant
3$.

Suppose $|S|>2$. We assume that $I=[s_{i},\dots,s_{n}]$ and
$J=[s_{1},\dots,s_{j}]$, with $j\leqslant n-1$, and we will show that
$i\geqslant j+2$.

Note that for each element $g\in W_{S}$ with its reduced expression
$g=s_{i_{1}}s_{i_{2}}\cdots s_{i_{k}}$, we define $\bar{g}=s_{i_{k}}\cdots
s_{i_{2}}s_{i_{1}}$. Note that $\bar{g}\in W_{S}$ does not depend on the
choice of reduced expression of $g$. Then $uv\leqslant_{L}\delta$ if and only
if $\bar{v}\bar{u}\leqslant_{R}\bar{\delta}$ if and only if
$\bar{v}\bar{u}\leqslant_{L}\bar{\delta}$. This allows us to reduce the $5-3$
case to the $3-5$ case, and the $4-3$ case to the $3-4$ case. Assume that $P$
is not of type $F_{4}$ with $i=2$ and $j=3$. Then without loss of generality,
up to this symmetry, we may assume that $J$ is of type $A_{j}$. Then we know
that $v=s_{1}s_{2}\cdots s_{j-1}s_{j}s_{j-1}\cdots s_{2}s_{1}$. We also have

$\displaystyle\delta=s_{1}\cdots s_{j}s_{j+1}\dots s_{n}=vs_{1}\cdots
s_{j-1}s_{j+1}\dots s_{n}$
$\displaystyle=(vs_{1}v^{-1})\cdots(vs_{j-1}v^{-1})(vs_{j+1}\dots
s_{n}v^{-1})v.$

By the Garside property, there exists $w\in[1,\delta]$ with reflection length
$n-1$ such that $w\cdot v=\delta$. Thus

$(vs_{1}v^{-1})\cdots(vs_{j-1}v^{-1})(vs_{j+1}v^{-1})\cdots(vs_{n}v^{-1})$

is a minimal reflection decomposition of $w$. By [McC15, Lemma 3.7],
$vs_{k}v^{-1}\leqslant_{L}w$ for $k\neq j$. Since $u\cdot
v\leqslant_{L}\delta$, we know that $u\cdot v\leqslant_{R}\delta$ by Theorem
2.16. Thus $u\leqslant_{R}w$.

According to [Bes03, Lemma 1.4.3], the element $w$ is a Garside element for
the parabolic subgroup $P_{w}$ of $P$ generated by the reflections which are
$\leqslant_{L}$-smaller than $w$. Since $w$ has reflection length $n-1$, this
subgroup $P_{w}$ equals

$P_{w}=\langle vs_{1}v^{-1},\cdots
vs_{j-1}v^{-1},vs_{j+1}v^{-1},\dots,vs_{n}v^{-1}\rangle=v\left(\langle
s_{1},\dots,s_{j-1}\rangle\times\langle
s_{j+1},\dots,s_{n}\rangle\right)v^{-1}.$

As $u\leqslant_{R}w$, we know that $u\in P_{w}$ by [BDSW14, Theorem 1.4].
Hence $v^{-1}uv\in W_{S\setminus\\{s_{j}\\}}$. By Lemma 2.1,
$\operatorname{Supp}(v^{-1}uv)\subset\\{s_{1},\ldots,s_{j-1}\\}$ or
$\operatorname{Supp}(v^{-1}uv)\subset\\{s_{j+1},\ldots,s_{n}\\}$. Thus $u\in
v\left(\langle s_{1},\ldots,s_{j-1}\rangle\right)v^{-1}$ or $u\in
v\left(\langle s_{j+1},\ldots,s_{n}\rangle\right)v^{-1}$. Also since

$v\left(\langle s_{1},\ldots,s_{j-1}\rangle\right)v^{-1}\subset
W_{\\{s_{1},\ldots,s_{j}\\}},$

we rule out that $u\in v\left(\langle
s_{1},\ldots,s_{j-1}\rangle\right)v^{-1}$, hence $u\in v\left(\langle
s_{j+1},\ldots,s_{n}\rangle\right)v^{-1}$. In particular,

$u\in v\left(\langle s_{j+1},\ldots,s_{n}\rangle\right)v^{-1}\cap
W_{I}=\langle vs_{j+1}v^{-1},s_{j+2},\ldots,s_{n}\rangle\cap\langle
s_{i},\ldots,s_{n}\rangle.$

Assume by contradiction that $i\leqslant j+1$. Let $P=\langle
vs_{j+1}v^{-1},s_{j+2},\ldots,s_{n}\rangle\cap\langle
s_{i},\ldots,s_{n}\rangle$. By [Qi07], $P$ is a parabolic subgroup of $W_{S}$.
Note that $P\supset W_{\\{s_{j+2},\ldots,s_{n}\\}}$. On the other hand,
$vs_{j+1}v^{-1}\notin W_{\\{s_{i},\ldots,s_{n}\\}}$ as
$vs_{j+1}v^{-1}=s_{1}s_{2}\cdots s_{j}s_{j+1}s_{j}\cdots s_{1}$ and the word
$s_{1}s_{2}\cdots s_{j}s_{j+1}s_{j}\cdots s_{1}$ is reduced by Lemma 5.3.
Hence $P=W_{\\{s_{j+2},\ldots,s_{n}\\}}$, contradicting that $i\leqslant j+1$.

The remaining case is in type $F_{4}$ with $i=2$ and $j=3$. Then $v$ is a
reflection inside the Coxeter group of $B_{3}$ generated by $s_{1},s_{2}$ and
$s_{3}$, which has Dynkin diagram $3-4$. Consider the canonical representation
of Coxeter group of type $B_{3}$ acting on $\mathbb{R}^{3}$. Then $s_{1}$ acts
by the orthogonal reflection along $x_{1}=x_{2}$, $s_{2}$ acts by the
orthogonal reflection along $x_{2}=x_{3}$, and $s_{3}$ acts by the orthogonal
reflection along $x_{3}=0$. Note that there are nine reflection in
$W_{s_{1},s_{2},s_{3}}$, whose reflection hyperplanes are $x_{i}=\pm x_{j}$
for $1\leq i\neq j\leq 3$ and $x_{i}=0$ for $1\leq i\leq 3$. Note that
reflections along $x_{i}=x_{j}$ for $1\leq i\neq j\leq 3$ are supported on
$W_{s_{1},s_{2}}$; reflections along $x_{2}=\pm x_{3}$ or $x_{i}=0$ for
$i=2,3$ are supported on $W_{s_{2},s_{3}}$. This gives 6 reflections in total.
The remaining three reflections in $W_{s_{1},s_{2},s_{3}}$ give all the
possibilities of $v$. More precisely, reflection along $x_{1}=0$ gives
$v=s_{1}s_{2}s_{3}s_{2}s_{1}$, reflection along $x_{1}+x_{3}=0$ gives
$v=s_{1}s_{3}s_{2}s_{3}s_{1}$, and reflection along $x_{1}+x_{2}=0$ gives
$v=s_{2}s_{3}s_{2}s_{1}s_{2}s_{3}s_{2}$.

The case $v=s_{1}s_{2}s_{3}s_{2}s_{1}$ is identical to before. Now we assume
$v=s_{1}s_{3}s_{2}s_{3}s_{1}=s_{3}s_{1}s_{2}s_{1}s_{3}=s_{3}s_{2}s_{1}s_{2}s_{3}$.
Then

$\displaystyle\delta=s_{1}s_{2}s_{3}s_{4}=s_{1}s_{2}s_{3}s_{4}s_{3}s_{2}s_{1}s_{2}s_{3}v$
$\displaystyle=(s_{1}s_{2}s_{3}s_{4}s_{3}s_{2}s_{1})(s_{2})(s_{3})v.$

By the same argument as before, we know $u\in\langle
s_{1}s_{2}s_{3}s_{4}s_{3}s_{2}s_{1},s_{2},s_{3}\rangle$. Let

$\displaystyle P=\langle
s_{1}s_{2}s_{3}s_{4}s_{3}s_{2}s_{1},s_{2},s_{3}\rangle=s_{1}s_{2}s_{3}\langle
s_{4},s_{1},s_{2}s_{3}s_{2}\rangle s_{3}s_{2}s_{1}=s_{1}s_{2}s_{3}s_{2}\langle
s_{4},s_{2}s_{1}s_{2},s_{3}\rangle s_{2}s_{3}s_{2}s_{1}$
$\displaystyle=s_{1}s_{2}s_{3}s_{2}\langle s_{4},s_{1}s_{2}s_{1},s_{3}\rangle
s_{2}s_{3}s_{2}s_{1}=s_{1}s_{2}s_{3}s_{2}s_{1}\langle s_{4},s_{2},s_{3}\rangle
s_{1}s_{2}s_{3}s_{2}s_{1}.$

In particular, $P$ is a parabolic subgroup. Note that $u\in P\cap
W_{s_{2},s_{3},s_{4}}$.

By [Qi07], $P\cap W_{s_{2},s_{3},s_{4}}$ is a parabolic subgroup of $W_{S}$.
Note that $\langle s_{2},s_{3}\rangle\subset P\cap W_{s_{2},s_{3},s_{4}}$.
Moreover, $s_{1}s_{2}s_{3}s_{4}s_{3}s_{2}s_{1}\in P\setminus
W_{s_{2},s_{3},s_{4}}$ as $s_{1}s_{2}s_{3}s_{4}s_{3}s_{2}s_{1}$ is a reduced
word by Lemma 5.3. Thus $P\cap W_{s_{2},s_{3},s_{4}}=\langle
s_{2},s_{3}\rangle$. Thus $s_{4}\notin\operatorname{Supp}(u)$, which is a
contradiction.

It remains to look at the case $v=s_{2}s_{3}s_{2}s_{1}s_{2}s_{3}s_{2}$. Then

$\displaystyle\delta=s_{1}s_{2}s_{3}s_{4}=s_{1}s_{2}s_{3}s_{4}s_{2}s_{3}s_{2}s_{1}s_{2}s_{3}s_{2}v$
$\displaystyle=(s_{1}s_{2}s_{3}s_{4}s_{3}s_{2}s_{1})(s_{1}s_{2}s_{3}s_{2}s_{3}s_{2}s_{1})(s_{2}s_{3}s_{2})v.$

Note that
$s_{2}s_{3}s_{2}=s_{1}s_{2}s_{3}(s_{3}s_{2}s_{1}s_{2}s_{3}s_{2}s_{1}s_{2}s_{3})s_{3}s_{2}s_{1}=s_{1}s_{2}s_{3}(s_{1}s_{2}s_{3}s_{2}s_{1})s_{3}s_{2}s_{1}$.
Thus by repeating the previous discussion, we know

$u\in s_{1}s_{2}s_{3}\langle s_{4},s_{2},s_{1}s_{2}s_{3}s_{2}s_{1}\rangle
s_{3}s_{2}s_{1}=s_{1}s_{2}s_{3}s_{1}s_{2}\langle s_{4},s_{1},s_{3}\rangle
s_{2}s_{1}s_{3}s_{2}s_{1}.$

As $u$ belongs to a parabolic subgroup which splits as a product and
$s_{4}\in\operatorname{Supp}(u)$, we argue as before to deduce that

$u\in s_{1}s_{2}s_{3}s_{1}s_{2}\langle s_{4},s_{3}\rangle
s_{2}s_{1}s_{3}s_{2}s_{1}:=P.$

Thus $u\in P\cap W_{s_{2},s_{3},s_{4}}$. Note that

$\displaystyle
s_{1}s_{2}s_{3}s_{1}s_{2}(s_{3})s_{2}s_{1}s_{3}s_{2}s_{1}=s_{1}s_{2}s_{1}(s_{3}s_{2}s_{3}s_{2}s_{3})s_{1}s_{2}s_{1}$
$\displaystyle=s_{2}s_{1}s_{2}(s_{2}s_{3}s_{2})s_{2}s_{1}s_{2}=s_{2}s_{1}s_{3}s_{1}s_{2}=s_{2}s_{3}s_{2}\in
P\cap W_{s_{2},s_{3},s_{4}}.$

Thus $P\cap W_{s_{2},s_{3},s_{4}}$ is a parabolic subgroup of rank $\geq 1$.
On the other hand,

$s_{1}s_{2}s_{3}s_{1}s_{2}(s_{4})s_{2}s_{1}s_{3}s_{2}s_{1}=s_{1}s_{2}s_{3}s_{4}s_{3}s_{2}s_{1}.$

As $s_{1}s_{2}s_{3}s_{4}s_{3}s_{2}s_{1}$ is a reduced word in $W_{S}$ by Lemma
5.3, it can not be contained in $W_{s_{2},s_{3},s_{4}}$. Thus $P\cap
W_{s_{2},s_{3},s_{4}}\subsetneq P$. It follows that $P\cap
W_{s_{2},s_{3},s_{4}}=\langle s_{2}s_{3}s_{2}\rangle$, hence
$s_{4}\notin\operatorname{Supp}(u)$, which is a contradiction. ∎

######  Corollary 5.5.

Under the same setting of Lemma 5.4, the conclusion of Lemma 5.4 holds for any
$u,v\in U$ such that $\operatorname{Supp}(u)$ and $\operatorname{Supp}(v)$ are
irreducible.

######  Proof.

Let $v=r_{1}r_{2}\cdots r_{k}$ with $r_{i}\in R_{U}$ be a minimal reflection
decomposition of $v$. Then there exists $1\leq i\leq k$ such that
$\min(I)-1\subset\operatorname{Supp}(r_{i})$. Suppose $I\subset J$ is not
true. Then
$\operatorname{max}(\operatorname{Supp}(r_{i}))<\operatorname{max}(\operatorname{Supp}(J))$.
Hence
$\operatorname{max}(\operatorname{Supp}(r_{i}))+1\in\operatorname{Supp}(J)$.
We write a minimal reflection decomposition of $u$ as $u=t_{1}t_{2}\cdots
t_{m}$. Then there exists $1\leq j\leq m$ such that
$\operatorname{max}(\operatorname{Supp}(r_{i}))+1\subset\operatorname{Supp}(t_{j})$.
As $u\cdot v\in\delta$, we know $r_{1}\cdots r_{k}t_{1}\cdots t_{m}$ is a
minimal reflection decomposition of $uv$. In particular $r_{1}\cdots
r_{k}t_{1}\cdots t_{m}\leq_{L}\delta$. By [McC15, Lemma 3.7], $r_{i}\cdot
t_{j}\leq_{L}\delta$. By construction we have
$\min(\operatorname{Supp}(t_{j}))-1\in\operatorname{Supp}(r_{i})$, and
$\operatorname{Supp}(r_{i})$ does not contain $\operatorname{Supp}(t_{j})$,
which contradicts Lemma 5.4. Thus the corollary is proved. ∎

###  5.2 Cyclic-type Artin groups

Let $W_{S}$ be a cyclic type Coxeter group (cf. Table 1). We take a cyclic
order on $S$ coming from its Dynkin diagram, and denote elements in $S$ as
elements in $\mathbb{Z}/n\mathbb{Z}$. For each $i\in\mathbb{Z}/n\mathbb{Z}$,
consider the dual Garside interval $U_{i}$ in $P_{S\char 92\relax i}$ with
respect to the dual Garside element $\delta_{i}=s_{i+1}s_{i+2}\dots
s_{n}s_{1}\dots s_{i-1}$. Let $U=\cup_{i\in\mathbb{Z}/n\mathbb{Z}}U_{i}$. It
is clear that this set $U$ satisfies the assumptions of Theorem 4.3.

For each $i\in\mathbb{Z}/n\mathbb{Z}$, consider the set $R_{U_{i}}\subset
U_{i}$ of reflections in the spherical parabolic subgroup $P_{S\char 92\relax
i}$, and let $R_{U}=\cup_{i\in\mathbb{Z}/n\mathbb{Z}}R_{U_{i}}\subset U$.

######  Proposition 5.6.

The sets $R$ and $U$ satisfy all assumptions from Proposition 4.2. In
particular, if $A_{\Gamma}$ is of cyclic type, then
$A_{\Gamma}\times\mathbb{Z}$ is a Garside group.

######  Proof.

We verify each assumption of Proposition 4.2 as follows.

  1. 1.

Any $u\in U_{i}$ can be written as a product of elements in $R_{U_{i}}$ which
is a minimal length reflection factorization.

  2. 2.

Let $r_{1},\dots,r_{m}\in U_{R}$ such that $u=r_{1}\cdot r_{2}\dots\cdot
r_{m}\in U$. Let $i\in\mathbb{Z}/n\mathbb{Z}$ such that $u\in U_{i}$. Then
$r_{1},\dots,r_{n}\in P_{S\char 92\relax i}$ by [BDSW14, Theorem 1.4], and
$r_{1}\cdot r_{2}\dots\cdot r_{m}\leqslant_{L}\delta_{i}$. So both $r_{1}\dots
r_{n-1}$ and $r_{2}\dots r_{n}$ belong to $P_{S\char 92\relax i}$, and also
they are both prefixes of $\delta_{i}$. Hence $r_{1}\dots r_{n-1}$ and
$r_{2}\dots r_{n}$ belong to $U_{i}$.

  3. 3.

Let $r,r^{\prime}\in U_{R}$ which admit a common left upper bound $u\in U$.
Let $i\in\mathbb{Z}/n\mathbb{Z}$ such that $u\in U_{i}$. By [BDSW14, Theorem
1.4], $r,r^{\prime}\in U_{i}$ and $u$ is a common left upper bound for
$r,r^{\prime}$ in $(U_{i},\leq_{L})$. In particular,
$r,r^{\prime}\leqslant_{L}\delta_{i}$, so $r$ and $r^{\prime}$ admit a unique
left join $u_{i}$ in $(U_{i},\leq_{L})$. Now we show $u_{i}$ is also the join
of $r$ and $r^{\prime}$ in $(U,\leq_{L})$. Indeed, take an arbitrary left
upper bound $u^{\prime}$ of $r,r^{\prime}$ in $U$. Suppose $u^{\prime}\in
U_{j}$. Then as before we know $r,r^{\prime}\in U_{j}$ and $u^{\prime}$ is a
common left upper bound of $r,r^{\prime}$ in $(U_{j},\leq_{L})$. Let $u_{j}$
(resp. $u_{ij}$) be the left join of $r,r^{\prime}$ in $(U_{j},\leq_{L})$
(resp. $(U_{i}\cap U_{j},\leq_{L})$). One readily verifies that
$u_{ij}=u_{i}$, $u_{ij}=u_{j}$ and $u_{j}\leq_{L}u^{\prime}$. Thus
$u_{i}\leq_{L}u^{\prime}$, implying $u_{i}$ is the left join of $r$ and
$r^{\prime}$ in $U$. The case of common right upper bound is similar.

  4. 4.

Let $a,u,v,w$ be as in Proposition 4.2 (4). Let
$I,J,K\subset\mathbb{Z}/n\mathbb{Z}$ denote $\operatorname{Supp}(a)$,
$\operatorname{Supp}(u)$ and $\operatorname{Supp}(v)$ respectively. We first
prove $I\cup J\cup K\subsetneq\mathbb{Z}/n\mathbb{Z}$. Suppose by
contradiction that $I\cup J\cup K=\mathbb{Z}/n\mathbb{Z}$. Let
$\\{I_{i}\\}_{i=1}^{k}$ be the irreducible components of $I$. By Lemma 2.1,
$a=a_{1}\cdot a_{2}\cdot\ldots\cdot a_{k}$ such that
$\operatorname{Supp}(a_{i})=I_{i}$. As $a_{i}\cdot u\leq a\cdot u$, we know
$a_{i}\cdot u\in U$. Similarly, $a_{i}\cdot v\in U$ for $1\leq i\leq k$. As
$I\cup J\cup K=\mathbb{Z}/n\mathbb{Z}$, for each $a_{i}$, we know either
$\min(I_{i})-1\in J$ or $\min(I_{i})-1\in K$ (here $I_{i}$ inherits a linear
order from the cyclic order on $\mathbb{Z}/n\mathbb{Z}$, hence it make senses
to take about $\min(I_{i})$ and $\min(I_{i})-1$). If $\min(I_{j})-1\in J$, we
consider $I_{i}\cup K$, which is irreducible. As $a_{i}\cdots u\in U$, there
exists $i_{0}$ such that $a_{i}\cdot u\in U_{i_{0}}$. Then $I_{i}\cup K\subset
S\setminus\\{i_{0}\\}$ by [BDSW14, Theorem 1.4]. We endow
$S\setminus\\{i_{0}\\}$ with the linear order induced from the cyclic order on
$S$, then Corollary 5.7 implies $I_{i}\subset J$. This shows that
$I_{i}\subset J\cup K$ for each $i$. Thus $I\cup J\cup K=J\cup K$. However, as
$u$ and $v$ has a left join $w$ in $U$, there exists $i^{\prime}_{0}$ such
that $w\in U_{i^{\prime}_{0}}$, hence $u,v\in U_{i^{\prime}_{0}}$ by [BDSW14,
Theorem 1.4]. Thus $J\cup K\subsetneq S$, which is a contradiction.

Let $i\in S$ such that $I\cup J\cup K\in S\setminus\\{i\\}$. As $w$ is a left
join of $u$ and $v$ in $U$, by the discussion in item 3,
$\operatorname{Supp}(w)\in J\cup K$ and $w$ is the left join of $u$ and $v$ in
$(U_{i},\leq_{L})$. Thus
$\operatorname{Supp}(a)\cup\operatorname{Supp}(w)\subset U_{i}$. Clearly
$aw\in U_{i}\subset U$. It remains to show $|aw|=|a|+|w|$. As
$(U_{i},\leq_{L})$ is a lattice, $a\cdot u$ and $a\cdot v$ has a left join in
$U_{i}$, denoted by $a^{\prime}$. As $a\leq_{L}a^{\prime}$, we know
$a^{\prime}=a\cdot w^{\prime}$ for some $w^{\prime}\in U_{i}$. By cancellation
property in $U_{i}$, we know $u\leq_{L}w^{\prime}$ and $u\leq_{L}w^{\prime}$.
Thus $w\leq_{L}w^{\prime}$. Then $w^{\prime}=w\cdot w_{0}$ for $w_{0}\in
U_{i}$. Then $|a^{\prime}|=|a|+|w^{\prime}|=|a|+|w|+|w_{0}|$. As
$a^{\prime}=aww_{0}$, $|a^{\prime}|\leq|aw|+|w_{0}|$. Thus $|aw|=|a|+|w|$.

  5. 5.

This is similar to the previous item.

  6. 6.

Let $a,b,u,v,x$ be as in Proposition 4.2 (6). Let
$I_{a}=\operatorname{Supp}(a)$. Similarly we define $I_{b},I_{u}$ and $I_{v}$.
We claim either $I_{a}\cup I_{b}\subsetneq S$ or $I_{u}\cup I_{v}\subsetneq
S$. Assume by contradiction that $I_{a}\cup I_{b}=S$ and $I_{u}\cup I_{v}=S$.
As $a\cdot x\cdot u\in U$, there there exists $i\in S$ such that $a\cdot
x\cdot u\leq_{L}\delta_{i}$. By [McC15, Lemma 3.7], $a\cdot u\leq_{L}a\cdot
x\cdot u\leq_{L}\delta_{i}$, thus $a\cdot u\in U_{i}\subset U$. Similarly,
$a\cdot v,b\cdot u,b\cdot v\in U$. As $I_{a}\cup I_{b}=S$, either
$\min(I_{u})-1\subset I_{a}$ or $\min(I_{u})-1\subset I_{b}$. If
$\min(I_{u})-1\subset I_{a}$, as $a\cdot u\in U_{i}$, we know from Lemma 5.4
that $I_{u}\subset I_{a}$. As $I_{u}\cup I_{v}=S$, $I_{a}\cup I_{v}=S$, which
contradicts $a\cdot v\in U$. The case of $\min(I_{u})-1\subset I_{b}$ is
similar.

If $I_{a}\cup I_{b}\subsetneq S$, then there is $i\in S$ such that $a,b\in
U_{i}$. As $(U_{i},\leq_{R})$ is lattice, $a$ and $b$ have a right join in
$(U_{i},\leq_{R})$, which is also a right join of $a$ and $b$ in $U$ by the
argument in item 3. The case $I_{u}\cup I_{v}\subsetneq S$ is similar.

∎

######  Corollary 5.7.

Under the same setting of Lemma 5.4, the conclusion of Lemma 5.4 holds for any
$u,v\in U$ such that $\operatorname{Supp}(u)$ and $\operatorname{Supp}(v)$ are
irreducible.

######  Proof.

Let $v=r_{1}r_{2}\cdots r_{k}$ with $r_{i}\in R_{U}$ be a minimal reflection
decomposition of $v$. Then there exists $1\leq i\leq k$ such that
$\min(I)-1\subset\operatorname{Supp}(r_{i})$. Suppose $I\subset J$ is not
true. Then
$\operatorname{max}(\operatorname{Supp}(r_{i}))<\operatorname{max}(\operatorname{Supp}(J))$.
Hence
$\operatorname{max}(\operatorname{Supp}(r_{i}))+1\in\operatorname{Supp}(J)$.
We write a minimal reflection decomposition of $u$ as $u=t_{1}t_{2}\cdots
t_{m}$. Then there exists $1\leq j\leq m$ such that
$\operatorname{max}(\operatorname{Supp}(r_{i}))+1\subset\operatorname{Supp}(t_{j})$.
As $u\cdot v\in\delta$, we know $r_{1}\cdots r_{k}t_{1}\cdots t_{m}$ is a
minimal reflection decomposition of $uv$. In particular $r_{1}\cdots
r_{k}t_{1}\cdots t_{m}\leq_{L}\delta$. By [McC15, Lemma 3.7], $r_{i}\cdot
t_{j}\leq_{L}\delta$. By construction we have
$\min(\operatorname{Supp}(t_{j}))-1\in\operatorname{Supp}(r_{i})$, and
$\operatorname{Supp}(r_{i})$ does not contain $\operatorname{Supp}(t_{j})$,
which contradicts Lemma 5.4. Thus the corollary is proved. ∎

Now the following is a consequence of Proposition 5.6 and Corollary 4.4.

######  Corollary 5.8.

Assume that $A_{\Gamma}$ is of hyperbolic cyclic type. Then $A_{\Gamma}$
satisfies the $K(\pi,1)$ conjecture and has trivial center.

##  6 Combination of cyclic type and spherical type Artin groups

An edge of a Coxeter presentation graph is _large_ if it its label is $\geq
3$. For each induced subgraph $\Lambda\subset\Gamma$, we define
$\Lambda^{\perp}$ to be the induced subgraph spanned by vertices of $\Gamma$
which commute with every vertex in $\Lambda$.

We will be considering orientation of each large edges of $\Gamma$. For the
moment suppose $\Gamma$ is spherical and we orient each large edge. We say a
Coxeter element of $\Gamma$ is compatible with such orientation if whenever
there is an oriented edge from $s_{1}\in S$ to $s_{2}\in S$, then $s_{1}$
appears before $s_{2}$ in the expression of the Coxeter element.

######  Lemma 6.1.

Given a spherical Coxeter presentation graph $\Gamma$ with orientation on its
large edges, any two Coxeter elements that are compatible with the orientation
are equal.

######  Proof.

We will prove it by induction on the rank of $\Gamma$.

Let us assume that $w=s_{1}\dots s_{n}$ and $w^{\prime}=s^{\prime}_{1}\dots
s^{\prime}_{n}$ are two Coxeter elements that are compatible with the
orientation. Let $i\in\\{1,\dots,n\\}$ such that $s^{\prime}_{i}=s_{1}$.
Assume that, among all possible reduced expressions of $w^{\prime}$ that are
compatible with the orientation, the position $i$ of $s_{1}$ is minimal. We
will prove that $i=1$.

Assume by contradiction that $i>1$. Since $i$ is minimal, we deduce that the
edge between $s^{\prime}_{i-1}$ and $s^{\prime}_{i}$ has label $\geq 3$. As
$w^{\prime}$ is compatible with the orientation, we deduce that the edge
between $s^{\prime}_{i-1}$ and $s^{\prime}_{i}$ is oriented from
$s^{\prime}_{i-1}$ to $s^{\prime}_{i}$. As $w$ is also compatible with the
orientation and $s^{\prime}_{i}=s_{1}$, we deduce that this edge is oriented
from $s^{\prime}_{i}$ to $s^{\prime}_{i-1}$. This is a contradiction.

So $s^{\prime}_{1}=s_{1}$. By induction, we deduce that $s_{2}\dots
s_{n}=s^{\prime}_{2}\dots s^{\prime}_{n}$, hence $w=w^{\prime}$. ∎

######  Lemma 6.2.

Given a spherical Coxeter presentation graph $\Gamma$ with orientation on its
large edges and let $\delta$ be the Coxeter element which is compatible with
the orientation. Let $[1,\delta]$ be the collection of elements in
$W_{\Gamma}$ that are prefixes of $\delta$ with respect to the reflection
length on $W_{\Gamma}$.

Give two reflections $r_{1},r_{2}\in[1,\delta]$ such that
$r_{1}r_{2}\in[1,\delta]$. Take
$s_{1}\in\operatorname{Supp}(r_{1})\setminus\operatorname{Supp}(r_{2})$ and
$s_{2}\in\operatorname{Supp}(r_{2})\setminus\operatorname{Supp}(r_{1})$. Then
either $s_{1}$ and $s_{2}$ commute, or there is an oriented edge from $s_{1}$
to $s_{2}$.

######  Proof.

We argue by contradiction and assume there is an oriented edge from $s_{2}$ to
$s_{1}$. Let $\Lambda$ be the Dynkin diagram, which is a tree. Then we cut
$\Lambda$ along the midpoint of the edge $\overline{s_{2}s_{1}}$ into two
subtrees with $s_{i}\in\Lambda_{i}$ for $i=1,2$. By Lemma 2.1,
$\operatorname{Supp}(s_{i})\subset\Lambda_{i}$ for $i=1,2$. The edge
orientation on $\Lambda$ induces edge orientation on $\Lambda_{i}$ for
$i=1,2$. Let $\delta_{i}$ be the Coxeter element in $A_{\Lambda_{i}}$ which is
compactible with the edge orientation on $\Lambda_{i}$. As vertices in
$\Lambda_{1}\setminus\\{s_{1}\\}$ commute with vertices in
$\Lambda_{2}\setminus\\{s_{2}\\}$, Lemma 6.1 implies that
$\delta=\delta_{2}\delta_{1}$. As $r_{i}$ is a reflection in
$A_{\Lambda_{i}}$, we know $r_{1}\leqslant_{L}\delta_{1}$ and
$r_{2}\leqslant_{R}\delta_{2}$ by [Bes03, Lemma 1.3.3]. In particular
$\delta_{1}$ has a minimal reflection decomposition of form
$\delta_{1}=r_{1}\cdot r^{\prime}_{1}\cdot r^{\prime}_{2}\cdots
r^{\prime}_{k}$, and $\delta_{2}$ has a minimal reflection decomposition of
form $\delta_{2}=r^{\prime\prime}_{1}\cdots r^{\prime\prime}_{m}\cdot r_{2}$.
Thus

$r^{\prime\prime}_{1}\cdots r^{\prime\prime}_{m}\cdot r_{2}\cdot r_{1}\cdot
r^{\prime}_{1}\cdot r^{\prime}_{2}\cdots r^{\prime}_{k}=\delta.$

By [McC15, Lemma 3.7], $r_{2}\cdot r_{1}\leqslant_{L}\delta$. Thus $r_{2}\cdot
r_{1},r_{2}\cdot r_{2}\in[1,\delta]$, and these two elements are both common
upper bound for $r_{1}$ and $r_{2}$ with respect to $\leq_{L}$. Then
$r_{2}r_{1}=r_{1}r_{2}$ as $([1,\delta],\leq_{L})$ is a lattice. We write
$r_{i}$ as an reducible word $w_{i}$ in $W_{S}$. Then $w_{i}$ only uses from
letters from $\Lambda_{i}$, and $w_{1}w_{2},w_{2}w_{1}$ are reduced words.
Then by Tits’s solution to the word problem of Coxeter group, we know that it
is possible to apply the relators finitely many times to transform
$w_{1}w_{2}$ into $w_{2}w_{1}$. However, as $s_{2}$ is on the right side of
$s_{1}$ in $w_{1}w_{2}$ and $m(s_{1},s_{2})\geq 3$, and the property of having
at least one $s_{2}$ on the right side of $s_{1}$ is preserved under applying
the relations, this leads to a contradiction. ∎

Given a 4-cycle $\omega\subset\Gamma$ with consecutive vertices
$\\{x_{i}\\}_{i=1}^{4}$, a pair of antipodal vertices in $\omega$ means either
the pair $\\{x_{1},x_{3}\\}$, or the pair $\\{x_{2},x_{4}\\}$. A 4-cycle in
$\Gamma$ has _diagonal_ means it has a pair of antipodal vertices of $\omega$
which are connected by an edge in $\Gamma$.

######  Theorem 6.3.

Let $\Gamma$ be a Coxeter presentation graph such that

  * $\bullet$

each complete subgraph of $\Gamma$ is a join of a cyclic type graph and a
spherical type graph (we allow one of the join factors to be empty);

  * $\bullet$

for any cyclic type induced subgraph $\Lambda\subset\Gamma$, $\Lambda^{\perp}$
is spherical.

We assume in addition that there exists an orientation of all large edges of
$\Gamma$ such that

  * $\bullet$

the orientation restricted to each cyclic type subgraph of $\Gamma$ gives a
consistent orientation on the associated circle;

  * $\bullet$

if $\omega$ is a 4-cycle in $\Gamma$ with a pair of antipodal points $x_{1}$
and $x_{2}$ such that each edge of $\omega$ containing
$x_{i}\in\\{x_{1},x_{2}\\}$ is either not large or oriented towards $x_{i}$,
then the cycle has a diagonal.

Then $A_{\Gamma}\times\mathbb{Z}$ is a Garside group.

######  Proof.

Let $S$ be the vertex set of $\Gamma$. Let $I\subset S$ be a spherical subset.
We define $\delta_{I}$ be a product of all elements in $I$ in an order which
is compatible with the orientation of $\Gamma$ in sense explained before the
lemma. Then $\delta_{I}$ is well-defined by Lemma 6.1.

We also view $\delta_{I}$ as an element in the Coxeter group $W_{\Gamma}$. Let
$\mathcal{S}$ be the collection of all spherical subset of $S$. Define
$U=\cup_{I\in\mathcal{S}}[1,\delta_{I}]$, where $[1,\delta_{I}]$ denotes the
interval in $W_{\Gamma}$ with respect to the reflection length. It is clear
that $U$ satisfies the assumptions of Theorem 4.3.

We now verify that $U$ satisfies all the requirements in Proposition 4.2.

By [BDSW14, Theorem 1.4], any minimal length reflection decomposition of an
element $a\in[1,\delta_{I}]$ only involves reflections in $W_{I}$. On the
other hand, by [Bes03, Lemma 1.3.3], for any reflection $r\in W_{I}$, there
exists a minimal length reflection decomposition of $\delta_{I}$ starting with
$r$, thus $r\in U$. Now Assumptions 1 and 2 of Proposition 4.2 follows.

For Assumption 3 of Proposition 4.2, if $r_{1},r_{2}\in R_{U}$ has an left
common upper bound $a\in U$, then there exists a spherical subset $I\in S$
such that $a\in[1,\delta_{I}]$. By [BDSW14, Theorem 1.4], $r_{1},r_{2}\in
W_{I}$. As in the previous paragraph, we know $r_{1},r_{2}\in[1,\delta_{I}]$.
As $([1,\delta_{I}],\leqslant_{L})$ is a lattice, we know $r_{1}$ and $r_{2}$
has left join $a$ in $([1,\delta_{I}],\leqslant_{L})$. By the same argument as
in the verification of Assumption 3 in Proposition Proof, we know $a$ is also
the left join of $r_{1}$ and $r_{2}$ in $(U,\leqslant_{L})$.

Now we verify Assumption 4 of Proposition 4.2.

For any $w\in W_{\Gamma}$, let $I_{w}=\operatorname{Supp}(w)$. We claim that
if $a,b\in U$ and $a\cdot b\in U$ (recall that $a\cdot b$ means $|ab|=|a|+|b|$
with $|\cdot|$ denotes the reflection length), then $I_{ab}=I_{a}\cup I_{b}$.
Note that $I_{ab}\subset I_{a}\cup I_{b}$ is clear. Now let
$a=r_{1}r_{2}\cdots r_{n}$ and $b=r^{\prime}_{1}r^{\prime}_{2}\cdots
r^{\prime}_{m}$ be minimal length reflection decomposition of $a$ and $b$. By
[BDSW14, Theorem 1.4],
$\operatorname{Supp}(r_{i})\subset\operatorname{Supp}(a)$ for each $i$, thus
$\operatorname{Supp}(a)=\cup_{i=1}^{n}\operatorname{Supp}(r_{i})$. Similarly
$\operatorname{Supp}(b)=\cup_{i=1}^{m}\operatorname{Supp}(r^{\prime}_{i})$. As
$a\cdot b\in U$, $r_{1}\cdots r_{n}r^{\prime}_{1}\cdots r^{\prime}_{m}$ is a
minimal length reflection decomposition of $a$ and $b$. As $ab\in A_{I_{ab}}$,
we know from [BDSW14, Theorem 1.4] that $r_{i},r^{\prime}_{i}\in A_{I_{ab}}$.
Similarly as before
$\operatorname{Supp}(ab)=(\cup_{i=1}^{n}\operatorname{Supp}(r_{i}))\cup(\cup_{i=1}^{m}\operatorname{Supp}(r^{\prime}_{i}))$.
Thus $I_{a}\subset I_{ab}$ and $I_{b}\subset I_{ab}$. Now the claim follows.

Let $a,b,u,w$ be as in Assumption 4 of Proposition 4.2. Then $I_{a}\cup
I_{u}=I_{au}$, which spans a complete subgraph of $\Gamma$. Similarly,
$I_{a}\cup I_{v}$ spans a complete subgraph of $\Gamma$. By the previous
paragraph, if $u\leq_{L}w$ and $v\leq_{L}w$, then $I_{u}\subset I_{w}$ and
$I_{v}\subset I_{w}$. Hence $I_{v}\cup I_{w}$ spans a complete subgraph of
$\Gamma$. Thus $I=I_{v}\cup I_{u}\cup I_{a}$ spans a complete subgraph of
$\Gamma$. Then $I=I_{1}\cup I_{2}$ where $I_{1}$ is a cyclic type irreducible
component of $I$ and $I_{2}$ is the union of all irreducible spherical
components of $I$. By Lemma 2.1, $a=a_{1}\cdot a_{2}$ for $a_{i}\in
W_{I_{i}}\cap U$ for $i=1,2$, $u$ belongs to either $W_{I_{1}}$ or
$W_{I_{2}}$, and $v$ belongs to either $W_{I_{1}}$ or $W_{I_{2}}$. If $u,v\in
W_{I_{1}}$, then Proposition Proof implies that $a_{1}\cdot w\in U\cap
W_{I_{1}}$, hence $a\cdot w=a_{2}\cdot a_{2}\cdot w\in U$. If exactly one of
$\\{u,v\\}$, say $u$, is in $W_{I_{1}}$, then $w=u\cdot v$, hence $a\cdot
w=(a\cdot u)\cdot v\in U$. If each of $u,v$ is in $W_{I_{2}}$, then
$a_{2}\cdot w\in[1,\delta_{I_{2}}]$ as $([1,\delta_{I_{2}}],\leqslant_{L})$ is
a lattice. Thus $a\cdot w=a_{1}\cdot(a_{2}\cdot w)\in U$.

Assumption 5 of Proposition 4.2 can be verified similarly.

Now we verify Assumption 6 of Proposition 4.2. Let $a,b,u,v,x$ be as in
Assumption 6. As $a\cdot x\cdot u\in U$, by previous discussion we know that
$I_{a}\cup I_{x}\cup I_{u}=I_{axu}$. Thus $I_{a}\cup I_{u}$ span a complete
subgraph. Similarly, $I_{a}\cup I_{v}$, $I_{b}\cup I_{u}$, $I_{b}\cup I_{v}$
span complete graphs of $\Gamma$.

First we consider the case when $I_{a}\cup I_{b}$ spans a complete subgraph.
If $I_{a}\cup I_{b}$ is spherical, then $a$ and $b$ have a right join in
$([1,\delta_{I_{a}\cup I_{b}}],\leqslant_{L})$, hence in $(U,\leqslant_{L})$.
Now suppose $I_{a}\cup I_{b}$ spans a cyclic type subgraph of $\Gamma$. Note
that $I_{a}\cup I_{b}\cup I_{u}$ spans a complete subgraph of $\Gamma$. As
$I_{u}$ is irreducible by Lemma 2.1, thus either $I_{u}\subset I_{a}\cup
I_{b}$, or $I_{u}\subset(I_{a}\cup I_{b})^{\perp}$ by our assumption on
complete subgraphs of $\Gamma$. Similarly, either $I_{v}\subset I_{a}\cup
I_{b}$, or $I_{v}\subset(I_{a}\cup I_{b})^{\perp}$. If both $I_{u}\subset
I_{a}\cup I_{b}$ and $I_{v}\subset I_{a}\cup I_{b}$ hold, then $a\cdot
u,b\cdot u,a\cdot v,b\cdot v\in U\cap W_{I_{a}\cup I_{b}}$ by [BDSW14, Theorem
1.4] and we are reduced to Theorem Proof. If at least one of the two
statements $I_{u}\subset I_{a}\cup I_{b}$ and $I_{v}\subset I_{a}\cup I_{b}$
is false, then $I_{u}\cup I_{v}$ is spherical, which implies that $u$ and $v$
have a left join in $U$.

The case when $I_{u}\cup I_{v}$ spans a complete subgraph is similar. It
remains to consider the case that $I_{u}\cup I_{v}$ does not span a complete
subgraph of $\Gamma$, and $I_{a}\cup I_{b}$ does not span a complete subgraph
of $\Gamma$. Now we will show this remaining case actually does not exist,
hence finishes the proof.

Suppose $I_{u}\cup I_{v}$ is not complete. Take $s_{u}\in I_{u}$ and $s_{v}\in
I_{v}$ such that they are not adjacent in $\Gamma$. We hope to show $I_{a}\cup
I_{b}$ spans a complete subgraph of $\Gamma$. Take $s\in I_{a}$ and $t\in
I_{b}$. If $s\in I_{u}$, then $s$ and $t$ are adjacent as $I_{u}\cup I_{b}$
spans a complete subgraph. Now we assume $s\notin I_{u}$. Note that
$s_{u}\notin I_{a}$, otherwise $s_{u}$ and $s_{v}$ are adjacent in $\Gamma$.
As $a\cdot x\cdot u\in U$, we know $a\cdot u\in U$ by [McC15, Lemma 3.7]. Now
by Lemma 6.2, either $s$ and $s_{u}$ commute, or there is an oriented edge
from $s$ to $s_{u}$. Similarly, we know this sentence is still true if we
replace the ordered pair $(s,s_{u})$ in the statement by $(t,s_{u})$,
$(s,s_{v})$ and $(t,s_{v})$. Thus by our assumption, the 4-cycle
$s,s_{u},t,s_{v}$ in $\Gamma$ must have a diagonal. The diagonal must connect
$s$ and $t$, as $s_{u}$ and $s_{v}$ are not adjacent. Thus $I_{a}\cup I_{b}$
spans a complete subgraph of $\Gamma$. ∎

The following is a consequence of Theorem 6.3 and Corollary 4.5.

######  Corollary 6.4.

Let $\Gamma$ be a Coxeter presentation graph with vertex set $S$ satisfying
all the assumptions in Theorem 6.3. For each spherical $T\subset S$, we choose
a Coxeter element $w_{T}\in W_{T}$ compatible with the orientation of
$\Gamma$. Let

$U=\bigcup_{T\subset S\mbox{ spherical}}[e,w_{T}].$

Let $\widehat{U}$ be the lift of $U$ from $W_{S}$ to $A_{S}$ via the
isomorphism between the dual Artin group associated with $A_{T}$ for each
$T\subset S$ spherical and $A_{T}$ ([Bes03, Theorem 2.2.5]).

Let $X_{S}$ be the flag complex of the Cayley graph of $A_{S}$ with generating
set $\widehat{U}$. Then $X_{S}$ admits an $A_{S}$-equivariant CUB metric such
that each simplex of $X_{S}$ is equipped with a polyhedral norm as in [Hae22].

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