Cubicallike geometry of quasimedian graphs and applications to geometric group theory
Abstract
The class of quasimedian graphs is a generalisation of median graphs, or equivalently of CAT(0) cube complexes. The purpose of this thesis is to introduce these graphs in geometric group theory. In the first part of our work, we extend the definition of hyperplanes from CAT(0) cube complexes, and we show that the geometry of a quasimedian graph essentially reduces to the combinatorics of its hyperplanes. In the second part, we exploit the specific structure of the hyperplanes to state combination results. The main idea is that if a group acts in a suitable way on a quasimedian graph so that cliquestabilisers satisfy some nonpositively curved property \mathcal{P}, then the whole group must satisfy \mathcal{P} as well. The properties we are interested in are mainly (relative) hyperbolicity, (equivariant) \ell^{p}compressions, CAT(0)ness and cubicality. In the third part, we apply our general criteria to several classes of groups, including graph products, Guba and Sapir’s diagram products, some wreath products, and some graphs of groups. Graph products are our most natural examples, where the link between the group and its quasimedian graph is particularly strong and explicit; in particular, we are able to determine precisely when a graph product is relatively hyperbolic.
Résumé

La classe des graphes quasimédians est une généralisation des graphes médians, ou de manière équivalente, des complexes cubiques CAT(0). L’objectif de cette thèse est d’introduire ces graphes dans le monde de la théorie géométrique des groupes. Dans un premier temps, nous étendons la notion d’hyperplan définie dans les complexes cubiques CAT(0), et nous montrons que la géométrie d’un graphe quasimédian se réduit essentiellement à la combinatoire de ses hyperplans. Dans la deuxième partie de notre texte, qui est le cœur de la thèse, nous exploitons la structure particulière des hyperplans pour démontrer des résultats de combinaison. L’idée principale est que si un groupe agit d’une bonne manière sur un graphe quasimédian de sorte que les stabilisateurs de cliques satisfont une certaine propriété \mathcal{P} de courbure négative ou nulle, alors le groupe tout entier doit satisfaire \mathcal{P} également. Les propriétés que nous considérons incluent : l’hyperbolicité (éventuellement relative), les compressions \ell^{p} (équivariantes), la géométrie CAT(0) et la géométrie cubique. Finalement, la troisième et dernière partie de la thèse est consacrée à l’application des critères généraux démontrés précédemment à certaines classes de groupes particulières, incluant les produits graphés, les groupes de diagrammes introduits par Guba et Sapir, certains produits en couronne (permutationnels), et certains graphes de groupes. Les produits graphés constituent notre application la plus naturelle, où le lien entre le groupe et son graphe quasimédian associé est particulièrement fort et explicite; en particulier, nous sommes capables de déterminer précisément quand un produit graphé est relativement hyperbolique.
Contents:
 1 Introduction
 2 Cubicallike geometry of quasimedian graphs
 3 Metrizing quasimedian graphs
 4 Cubulating quasimedian graphs

5 Topical actions on quasimedian graphs I
 5.1 Creating invariant coherent systems I
 5.2 Topical and topicaltransitive actions
 5.3 Relatively hyperbolic groups acting on quasimedian graphs
 5.4 Cubulating groups acting on quasimedian graphs I
 5.5 ATmenability and aBmenability
 5.6 Creating coherent invariant systems II
 5.7 Equivariant \ell^{p}compression
 6 Inflating the hyperplanes of a quasimedian graph
 7 Topical actions on quasimedian graphs II
 8 Application to graph products
 9 Application to wreath products
 10 Application to diagram products
 11 Application to rightangled graphs of groups
 12 Open problems
1 Introduction
CAT(0) cube complexes were introduced by Gromov in his seminal paper [Gro87] as a convenient source of examples of CAT(0) and CAT(1) groups. But their strength really appeared with the recognition of the central role played by the combinatorics of their hyperplanes, initiated by Sageev in his thesis [Sag95]. Since then, several still open conjectures for CAT(0) spaces were verified for CAT(0) cube complexes, including the (bi)automaticity of cubulated groups [NR98a], the Tits Alternative for groups acting freely on finitedimensional CAT(0) cube complexes [SW05], and the Rank Rigidity Conjecture [CS11]. Recently, CAT(0) cube complexes were also crucial in the proof of the famous virtual Haken conjecture [Ago13].
Independently, Chepoï [Che00] and Roller [Rol98] realised that the class of CAT(0) cube complexes can be naturally identified with the class of the socalled median graphs. These graphs were known by graph theoretists for a long time since they were introduced by Nebeský in 1971 [Neb71]. Since then, several classes of graphs were introduced as generalisations of median graphs (see for instance [CCHO14] and references therein), including the main subjet of this article, quasimedian graphs, which were introduced by Mulder in 1980 [Mul80] and more extensively studied by Bandelt, Mulder and Wilkeit in 1994 [BMW94]. The ambition of this article is to introduce quasimedian graphs into the world of geometric group theory as natural and powerful objects.
The first part of our work consists in extending the definition of hyperplanes from CAT(0) cube complexes to quasimedian graphs, and then showing that the geometry of quasimedian graphs essentially reduces to the combinatorics of their hyperplanes. The results we prove are quite similar to those which hold for CAT(0) cube complexes. The main difference is that cutting along a hyperplane in a quasimedian graph may produce arbitrarily many connected components (but always at least two), whereas cutting along a hyperplane in a CAT(0) cube complex produces exactly two connected components. In fact, the analogy between quasimedian graphs and CAT(0) cube complexes is so strong that a precise dictionnary is possible, allowing to translate any statement which holds for CAT(0) cube complexes into a statement which is expected to hold for quasimedian graphs; Table 1 is an attempt for such a dictionnary. For instance, we prove:
Proposition 1.1.
Let X be a quasimedian graph.

•
A hyperplane separates X into at least two connected components, called sectors; they are gated subgraphs.

•
The carrier of a hyperplane decomposes as a Cartesian product of a clique and a quasimedian subgraph.

•
For every finite collection of pairwise transverse hyperplanes, there exists a prism whose dual hyperplanes are precisely those hyperplanes.

•
A path in X is a geodesic if and only if it intersects each hyperplane at most once; as a consequence, the distance between two vertices is equal to the number of hyperplanes separating them.

•
If a vertex x does not belong to a gated subgraph C, then there exists a hyperplane separating x from C.
Therefore, if one keeps CAT(0) cube complexes in mind, we do not prove any surprising result, but this is precisely this similarity which turns out to be surprising. In fact, several results we prove are already known by graph theorists, possibly in a different language. The main objective of the first part of our work is to introduce a common formalism as the foundation of the rest of our work, but also of our current knowledge about quasimedian graphs.
CAT(0) cube complexes / Median graphs  Quasimedian graphs 

Hyperplanes  Hyperplanes [2.14] 
Halfspaces  Sectors [2.17, 2.19] 
Combinatorially convex subcomplexes  Gated subgraphs [2.2] 
Finite subcomplexes  Cubically finite subgraphs [2.70, 2.81] 
Cubes  Prisms [2.72] 
Median vertex  Quasimedian triangle [2.82] 
Pocsets  Popsets [2.41] 
Spaces with walls  Spaces with partitions [2.59] 
Once quasimedian graphs are studied, our purpose is to find information on groups acting on them. However, by considering pairs \{S,S^{c}\} where S is a sector, quasimedian graphs are naturally spaces with walls, so that a CAT(0) cube complex is associated to any quasimedian graph. As a consequence, admitting a “nice” action (eg. a geometric action) on a quasimedian graph produces a similar action on a CAT(0) cube complex (see Proposition 4.16 for a precise statement). So why do we care about quasimedian graphs? The first reason is that their geometries may be easier to handle than those of the corresponding CAT(0) cube complexes, allowing us to exploit the action further in order to deduce interesting properties of the group. Graph products are the typical examples where such a situation happens; more details will be given below. The second reason is that “suitable” actions on quasimedian graphs, which are far from being proper, lead to combination theorems. More precisely, given a group G acting on some quasimedian graph X, our strategy is the following:
 Step 1.

Fix a convenient set \mathcal{C} of representatives of cliques of X.
 Step 2.

For every C\in\mathcal{C}, use the action \mathrm{stab}(C)\curvearrowright C to transfer structures from \mathrm{stab}(C) to C (eg. a metric). The most convenient situation is when this action is transitive and free, so that an orbit map provides a bijection. Typically, \mathcal{C} decomposes into two parts \mathcal{C}=\mathcal{C}_{1}\sqcup\mathcal{C}_{2}: in the first one, the actions we are interested in are indeed transitive and free; and in the second part, we have no control on the actions, but the cliques are finite. Therefore, we are able to define structures on the cliques of \mathcal{C}_{1} from their stabilisers, and usually we put trivial structures (eg. discrete metrics) on the cliques of \mathcal{C}_{2}; because they are finite, this does not cause any trouble.
 Step 3.

Next, if the action of G on X is “wellbehaved”, it is possible to extend our collection of structures defined on the cliques of \mathcal{C} to a system of structures, defined on each clique of X, which is compatible with the graph structure and which is Ginvariant. This is not always possible, so we need to define carefully what a “suitable action” means, but when the extension is possible, it is unique.
 Step 4.

From such a system, we define a global structure on X which extends the “local structures”, and study how the global structure inherits its properties from the local ones.
 Step 5.

Use the action of G on X endowed with the global structure to find information about G (eg. existence of a good action on some metric space).
We introduce \mathcal{C}transitive actions on quasimedian graphs in order to make the extension mentionned in Step 2 possible; and similarly, topical actions for the extension mentionned in Step 3.
Remark 1.2.
In fact, \mathcal{C}transitive actions are stronger than what it is described in Step 2, because we require some control on the finite cliques of \mathcal{C}_{2}. The first reason for this choice is that this generality is sufficient to cover all our applications, so we chose to give a presentation as simple as possible. As a consequence, the results proved in Section 5 hold in a framework which is slightly more general. The second reason is that with our stronger hypotheses, we get a canonical way to identify an arbitrary clique of X to a clique of our set of representatives \mathcal{C}, and this will be fundamental in a construction described below, we call inflating the hyperplanes.
The second part of our work develops this strategy. Typically, the results we obtain have the following form: given a property of groups \mathcal{P} and a group G acting topicallytransitively on a quasimedian graph, such that the action satisfies some convenient finiteness conditions (depending on \mathcal{P}), if cliquestabilisers satisfy \mathcal{P} then so does G. The structures we are interested in are mainly metrics and collections of walls, in order to study the geometry of a group and its cubical properties; but we also consider measured wallspaces, spaces with labelled partitions and Lipschitz maps to L^{p}spaces in order to study aTmenability, a\mathcal{B}menability and (equivariant) \ell^{p}compressions respectively. Thus, we proved criteria for

•
acting metrically properly on a CAT(0) cube complex (Proposition 5.22);

•
acting geometrically on a CAT(0) cube complex (Proposition 5.23);

•
being aTmenable (Proposition 5.25);

•
being a\mathcal{B}menable (Proposition 5.26);

•
bounding below \ell^{p}compressions (Proposition 5.37);

•
being relatively hyperbolic (Theorem 5.17).
However, our strategy does not work for many interesting properties, because in Step 2 we define structures on cliques from their stabilisers. But many properties cannot be read directly from the group, for instance being CAT(0). The trick is the following. Let G be a group acting on a quasimedian graph X. If a cliquestabiliser \mathrm{stab}(C) acts on another space Y_{C} (with a point of trivial stabiliser and such that the action \mathrm{stab}(C)\curvearrowright C is transitive and free), then we can identify C with a subset of Y_{C}. Next, we add to C the missing points of Y_{C}, and doing this process in a suitable way for every clique of X produces a new quasimedian graph Y on which G acts. Now, the cliques of Y are naturally identified with some spaces, so that they are naturally endowed with the corresponding structures. Finally, reproducing our strategy from Step 3 produces other combination results. We find criteria for:

•
acting properly discontinuously on a CAT(0) cube complex (Proposition 7.4);

•
acting geometrically and virtually specially on some CAT(0) cube complex (Proposition 7.5);

•
being CAT(0) (Theorem 7.7).
The third and last part of this work is dedicated to the application of the previous criteria to specific classes of groups, which we now describe.
Graph products.
Given a simplicial graph \Gamma and a collection of groups \mathcal{G}=\{G_{u}\mid u\in V(\Gamma)\} indexed by the vertices of \Gamma, Green [Gre90] defined the graph product \Gamma\mathcal{G} as the quotient
\left(\underset{u\in V(\Gamma)}{\ast}G_{u}\right)/\langle\langle[g,h]=1,g\in G% _{u},h\in G_{v}\ \text{if}\ (u,v)\in E(\Gamma)\rangle\rangle. 
For instance, the geometry and the combinatorics of graph products were studied in [Alo96, AD13, BM15, Bou97, JS01, HM95, HW99b, Mei96, Mei94, Kim12, Mei95, AM15]. In our work, we are interested in the Cayley graph X(\Gamma,\mathcal{G}) defined from the generating set \bigsqcup\mathcal{G}, which turns out to be a quasimedian graph. Moreover, thanks to the normal form proved by Green in her thesis, we understand precisely the geodesics of X(\Gamma,\mathcal{G}), making X(\Gamma,\mathcal{G}) a very convenient geometric model of \Gamma\mathcal{G}. It is worth noticing that, for graph products of finitely many finite groups, X(\Gamma,\mathcal{G}) is a Cayley graph obtained from a finite generating set, so that it turns out to be a complete geometric realisation of the graph product \Gamma\mathcal{G}. Although we deduce from this observation that \Gamma\mathcal{G} must act geometrically on some CAT(0) cube complex, the precise description of X(\Gamma,\mathcal{G}) allows us to prove easily that \Gamma\mathcal{G} is virtually cocompact special, that it embeds quasiisometrically into a product of \chi(\Gamma) trees (see Theorem 8.30), and also to determine exactly when it is hyperbolic, so that we are able to reprove [Mei96, Theorem A] which characterizes the hyperbolic graph products (of arbitrary groups). In fact, applying our general criterion on relative hyperbolicity allows us to determine when a graph product is hyperbolic relatively to its factors. Thanks to our description of X(\Gamma,\mathcal{G}), the argument can be improved further, by following our argument in [Gen16b, Theorem 5.24], in order to characterize precisely when a graph product is relatively hyperbolic. More precisely, we associate to any finite simplicial graph \Gamma and to any collection of groups \mathcal{G} labelled by V(\Gamma), a collection \mathfrak{I}(\Gamma,\mathcal{G}) of subgraphs of \Gamma. We emphasize the fact that, the construction of \mathfrak{I}(\Gamma,\mathcal{G}) is explicit, algorithmic, and does not depend heavily on the vertexgroups: we only need to know which groups of \mathcal{G} are finite, other information about the vertexgroups being unnecessary. Then
Theorem 1.3.
Let \Gamma be a finite graph and \mathcal{G} a collection of finitely generated groups indexed by V(\Gamma). The graph product \Gamma\mathcal{G} is relatively hyperbolic if and only if \mathfrak{I}(\Gamma,\mathcal{G})\neq\{\Gamma\}. If so, \Gamma\mathcal{G} is hyperbolic relatively to \{\Lambda\mathcal{G}\mid\Lambda\in\mathfrak{I}(\Gamma,\mathcal{G})\}.
Next, by applying our general criteria, we immediately find that the properties of

•
being CAT(0) (Theorem 8.20);

•
acting geometrically on a CAT(0) cube complex (Theorem 8.17);

•
acting geometrically and virtually specially on a CAT(0) cube complex (Theorem 8.17);
are stable under graph products along finite graphs; that the properties of

•
acting metrically properly on a CAT(0) cube complex (Proposition 8.18);

•
being aTmenable (Corollary 8.19);

•
being a\mathcal{B}menable (Corollary 8.19);
are stable under graph products along countable graphs; and finally that acting properly discontinuously on a CAT(0) cube complex is stable under arbitrary graph products (Theorem 8.16). By applying our criterion about \ell^{p}compressions, we are also able to reprove [AD13, Corollary 4.4]:
Theorem 1.4.
Let \Gamma be a finite simplicial graph and \mathcal{G} a collection of finitely generated groups indexed by V(\Gamma). For every p\geq 1,
\alpha^{*}_{p}(\Gamma\mathcal{G})\geq\min\left(\frac{1}{p},\min\limits_{G\in% \mathcal{G}}\alpha_{p}^{*}(G)\right). 
Some of the results we prove are already known, but the point is that we are able to include all of them in a unique point of view, the study of quasimedian graphs, and that we are sometimes able to simplify or even improve them. In our opinion, quasimedian geometry is the most natural way to study graph products.
Wreath products.
The wreath product of two groups G and H is defined as the semidirect product
G\wr H=\left(\bigoplus\limits_{h\in H}G\right)\rtimes H, 
where H acts on the direct sum by permuting the coordinates. Wreath products are important in geometric group theory because they lead to interesting counterexamples [Dyu00, dC06, Akh08, Dym10], but their geometries remain widely unknown appart from a few particular cases [EFW13]. Motivated by the cubulation of \mathbb{Z}\wr\mathbb{Z} as a diagram group (see [GS99, Example 10] and [Far03]), we associate to any action of H on a CAT(0) cube complex (containing a vertex of trivial stabiliser) a quasimedian graph \mathfrak{W}, called the graph of wreaths, on which G\wr H acts topicallytransitively. We refer to the introduction of Section 9 for a description of the idea of the construction. First of all, by applying our criterion producing proper actions on cube complexes, we are able to reprove [Cor13, Theorem 5.C.3]:
Theorem 1.5.
Acting properly discontinuously on a CAT(0) cube complex is stable under wreath products.
It is worth noticing that, contrary to Cornulier’s proof, we get an explicit construction of the CAT(0) cube complex on which the wreath product acts (see Remark 9.30). But the new results on wreath products we prove in this paper concern equivariant \ell^{p}compressions. The paper [AGS06] written by Arzhantseva, Guba and Sapir, where they show (in particular) that the Hilbert space compression of \mathbb{Z}\wr\mathbb{Z} belongs to the interval [1/2,3/4], motivated a lot of works on finding estimates on the \ell^{p}compressions of wreath products; see for instance [NP08, NP11, Li10, Tes11, ANP09]. In this paper, we adapt ideas from [CN05a], which were formulated for CAT(0) cube complexes, to quasimedian graphs, and in particular to the graph of wreaths \mathfrak{W} for the study of wreath products. Our main result is the following:
Theorem 1.6.
Let G,H be two finitely generated groups. Suppose that H acts on a CAT(0) cube complex X with an orbit map which has compression \alpha. Then
\alpha_{p}^{*}(G\wr H)\geq\alpha\cdot TS(X)\cdot\min\left(\frac{1}{p},\alpha_{% p}^{*}(G)\right). 
This result is not an immediate application of our general criterion, because wreath products may not be quasiisometrically embedded into the geometric models we construct. Understanding the distorsion of this embedding leads to the introduction of the constant TS(X), depending on the geometry of the CAT(0) cube complex X. By noticing that TS(X)=1 for any uniformly locally finite hyperbolic CAT(0) cube complex X (see Proposition 9.45), we deduce our main application:
Corollary 1.7.
Let H be a hyperbolic group acting geometrically on some CAT(0) cube complex. For every finitely generated group G and every p\geq 1,
\alpha_{p}^{*}(G\wr H)\geq\min\left(\frac{1}{p},\alpha_{p}^{*}(G)\right), 
with equality if H is non elementary and p\in[1,2].
Diagram products.
In [GS99], Guba and Sapir defined a diagram product D(\mathcal{P},\mathcal{G},w) as the fundamental group of a 2complex of groups associated to a semigroup presentation \mathcal{P}=\langle\Sigma\mid\mathcal{R}\rangle, a collection of groups \mathcal{G} indexed by the alphabet \Sigma, and a base word w\in\Sigma^{+}. By noticing that diagram products can be described as diagrams whose edges are labelled by a new alphabet \Sigma(\mathcal{G})=\{(s,g)\mid s\in\Sigma,g\in G_{s}\}, we generalize the construction due to Farley of CAT(0) cube complexes associated to diagram groups [Far03] in order to produce a quasimedian graph on which D(\mathcal{P},\mathcal{G},w) acts topicallytransitively. This allows us to apply our different combination results.
So let \mathcal{P}=\langle\Sigma\mid\mathcal{R}\rangle be a semigroup presentation, \mathcal{G} a collection of groups indexed by \Sigma and w\in\Sigma^{+} a base word.

•
If the groups of \mathcal{G} act properly on CAT(0) cube complexes, then so does the diagram product D(\mathcal{P},\mathcal{G},w).
Moreover, assuming that \mathcal{P} is a finite presentation,

•
if the groups of \mathcal{G} act metrically properly on CAT(0) cube complexes, then so does the diagram product D(\mathcal{P};\mathcal{G},w);

•
if the groups of \mathcal{G} are aTmenable, then so is the diagram product D(\mathcal{P},\mathcal{G},w);

•
if the groups of \mathcal{G} are a\mathcal{B}menable, then so is the diagram product D(\mathcal{P},\mathcal{G},w).
And finally, assuming that \mathcal{P} is a finite presentation and that the class [w]_{\mathcal{P}} of w modulo \mathcal{P} is finite,

•
if the groups of \mathcal{G} act geometrically on CAT(0) cube complexes, then so does the diagram product D(\mathcal{P},\mathcal{G},w);

•
if the groups of \mathcal{G} are CAT(0), then so is the diagram product D(\mathcal{P},\mathcal{G},w).
We are also able to estimate the \ell^{p}compressions of diagram products when the class of the base word is finite; we will refer to this class of diagram products as cocompact diagram products, because this assumption amounts to say that the the quasimedian graph we define contains only finitely many orbits of cliques.
Theorem 1.8.
Let \mathcal{P}=\langle\Sigma\mid\mathcal{R}\rangle be a semigroup presentation, \mathcal{G} a collection of groups indexed by \Sigma and w\in\Sigma^{+} a base word. Suppose that \mathcal{P} is a finite presentation, that [w]_{\mathcal{P}} is finite, and that the groups of \mathcal{G} are finitely generated. For every p\geq 1, the diagram product D(\mathcal{P},\mathcal{G},w) satifies
\alpha_{p}^{*}(D(\mathcal{P},\mathcal{G},w))\geq\min\left(\frac{1}{p},\min% \limits_{G\in\mathcal{G}}\alpha_{p}^{*}(G)\right). 
We also deduce from our general combination results a characterisation of hyperbolic cocompact diagram products. We suspect that our statement holds without the cocompactness assumption; see Question 12.24 and the related discussion.
Theorem 1.9.
Let \mathcal{P}=\langle\Sigma\mid\mathcal{R}\rangle be a semigroup presentation, \mathcal{G} a collection of finitely generated groups indexed by \Sigma, and w\in\Sigma^{+} a base word whose class [w]_{\mathcal{P}} is finite. The diagram product D(\mathcal{P},\mathcal{G},w) is hyperbolic if and only if, for every non empty words u,v\in\Sigma^{+} such that w is equal to uv modulo \mathcal{P}, at least one of the two diagram products D(\mathcal{P},\mathcal{G},u) and D(\mathcal{P},\mathcal{G},v) is finite.
Finally, in order to better understand the structure of diagram groups, we introduce and study another type of actions on quasimedian graphs, called rotative actions. Our main result shows that a group admitting such an action decomposes as a semidirect product with a normal factor which is a graph product (see Theorem 10.54 for a precise statement). This structure result is of independent interest, and applies to our other classes of groups acting on quasimedian graphs, but in these cases we do not get new information. Once applied to diagram products, we prove
Theorem 1.10.
Let \mathcal{P}=\langle\Sigma\mid\mathcal{R}\rangle be a semigroup presentation, \mathcal{G} a collection of finitely generated groups indexed by \Sigma, and w\in\Sigma^{+} a base word. The diagram product decomposes as
D(\mathcal{P},\mathcal{G},w)=\Gamma(\mathcal{P},\mathcal{G},w)\rtimes D(% \mathcal{P},w), 
where D(\mathcal{P},w) is the underlying diagram group and \Gamma(\mathcal{P},\mathcal{G},w) a graph product whose vertexgroups are elements of \mathcal{G}.
We refer to Section 10.6 for a precise description of the graph product \Gamma(\mathcal{P},\mathcal{G},w). As a consequence, when the underlying diagram group is trivial and when the groups of \mathcal{G} are either trivial or infinite cyclic, our theorem produces new examples of rightangled Artin groups which are also diagram groups; concrete examples are given at the end of Section 10.6. The general question of determining which rightangled Artin groups are diagram groups remains open.
Rightangled graph of groups.
Our last class of groups acting on quasimedian graphs generalises the graph products. A rightangled graph of groups is a graph of groups \mathfrak{G} whose vertices are graph products, whose edges are “subgraph products”, and whose monomorphisms are canonical embeddings (see Section 11 for a precise definition). Concrete examples are given in Section 11.4. We use the normal form of fundamental groupoids of graph of groups proved in [Hig76] to deduce that a natural (connected component of a) Cayley graph of the fundamental groupoid of \mathfrak{G} turns out to be a quasimedian graph. Thus, we make the fundamental group of \mathfrak{G} act on a quasimedian graph. In particular, if all the factors of this product are finite, we find that the fundamental group acts geometrically on a CAT(0) cube complex, and we are able to determine precisely when it is hyperbolic (see Proposition 11.41 for a precise statement). When some factors are infinite, we would like to show that the action is topicaltransitive and then apply our criteria. However, it turns out that our action may not be topicaltransitive. This is also a motivation for introducing this class of groups: having examples to test possible future extensions of our work. The obstruction for the action to be topicaltransitive is clearly identified: by concatenating monomorphisms of our graph of groups, every factor G arises with a collection of automorphisms \Phi(G), and our action is topicaltransitive precisely when \Phi(G) is reduced to \{\operatorname{Id}\} for every factor G. Loosely speaking, the dynamics of the action of the fundamental group of \mathfrak{G} on its associated quasimedian graph is related to the dynamics of collections of automorphisms on factors, and the topicaltransitive situation corresponds to the simplest case. We suspect that our results extend to the situation where the \Phi(G)’s are finite collections of finiteorder automorphisms.
Let \mathfrak{G} be a rightangled graph of groups such that \Phi(G)=\{\operatorname{Id}\} for every factor G. By applying our criteria, we prove that

•
if the factors act properly on CAT(0) cube complexes, then so does the fundamental group of \mathfrak{G};
assuming that the underlying abstract graph is locally finite and that the simplicial graphs defining the graph products are finite, we are able to deduce that:

•
if the factors act metrically properly on CAT(0) cube complexes, then so does the fundamental group of \mathfrak{G};

•
if the factors are aTmenable, then so is the fundamental group of \mathfrak{G};

•
if the factors are a\mathcal{B}menable, then so is the fundamental group of \mathfrak{G};
and finally, if the underlying abstract graph and the simplicial graphs defining the graph products are all finite, then

•
If the factors act geometrically on CAT(0) cube complexes, then so does the fundamental group of \mathfrak{G};

•
if the factors are CAT(0), then so is the fundamental group of \mathfrak{G}.
(All these statements are proved in Section 11.2.) Moreover, because the orbit map from the fundamental group of \mathfrak{G} to its associated quasimedian graph turns out to be a quasiisometric embedding, it is also possible to estimate the equivariant \ell^{p}compressions of the group. More precisely,
Theorem 1.11.
Let \mathfrak{G} be a rightangled graph of groups such that \Phi(G)=\{\operatorname{Id}\} for every factor G. Suppose that the underlying abstract graph and the simplicial graphs defining the graph products are all finite, and that the factors are finitely generated. For every p\geq 1, the fundamental group \mathfrak{F}_{\omega} of \mathfrak{G} satifies
\alpha_{p}^{*}(\mathfrak{F}_{\omega})\geq\min\left(\frac{1}{p},\min\limits_{% \text{$G$ factor}}\alpha_{p}^{*}(G)\right). 
Finally, we exploit the quasimedian geometry of the fundamental group of \mathfrak{G} in order to determine precisely when it is hyperbolic. The characterisation we obtain is the following (we refer to Section 11.1 for a definition of the link of a factor).
Theorem 1.12.
Let \mathfrak{G} be a rightangled graph of groups such that \Phi(G)=\{\operatorname{Id}\} for every factor G. Suppose that the factors are finitely generated, and that the underlying abstract graph and the simplicial graphs defining the graph products are all finite. The fundamental group of \mathfrak{G} is hyperbolic if and only if the following conditions are satisfied:

•
any simplicial graph defining one of our graph products is squarefree;

•
there do not exist a loop p in the abstract graph, based at some vertex v\in V, and two non adjacent vertexgroups G,H in the graph product G_{v}, such that \varphi_{p}^{G}(G)=G and \varphi_{p}^{H}(H)=H;

•
the link of every factor is finite;

•
the factors are hyperbolic.
Organization of this paper.
Section 2 contains the first part of our work, dedicated to the study of quasimedian graphs from hyperplanes. Section 3 and Section 4 describe how to perform the fourth step of the strategy described above for metrics and wallspaces respectively, while the second and third steps are described in Section 5, in which we also state and prove our first general criteria. Section 6 describes how to inflate the hyperplanes of a quasimedian graph, and we use this construction in Section 7 to find our second family of general criteria. The next four sections are dedicated to applications to graph products, wreath products, diagram groups, and rightangled graph of groups. We conclude this paper by some open problems and questions in Section 12.
Acknowledgement.
I am grateful to Jérémie Chalopin and Victor Chepoï, for a useful discussion on generalisations of median graphs, which lead me to state and prove the arguments I wrote in an early draft in the context of quasimedian graphs; of course to my advisor, Peter Haïssinsky, for all our discussions during the last three years and for all his comments on my manuscript; and to Vincent Guirardel and Michah Sageev for having accepted to refer my thesis, with a special thought for Vincent and all his useful comments. Finally, I would like to thank Goulnara Arzhantseva, Indira Chatterji, Yves Cornulier and Victor Chepoï for having accepted to participate to my jury.
2 Cubicallike geometry of quasimedian graphs
In this section, we introduce quasimedian graphs and study their geometries. The main idea we want to stress out is that one may define hyperplanes in quasimedian graphs so that arguments holding for CAT(0) cube complexes can be naturally translated to quasimedian graphs.
2.1 Quasimedian graphs
Quasimedian graphs are the main objects studied in this work. Among all the equivalent definitions of these graphs, it seems that the more convenient definition for our purposes is to see quasimedian graphs as particular weakly modular graphs. Weakly modular graphs were introduced in [Che89] and [BC96]; see also [CCHO14]. Quasimedian graphs appeared independently under different definitions in several places in the litterature; see [BMW94] and references therein for more information.
Definition 2.1.
A graph is weakly modular if it satisfies the following two conditions:
 (triangle condition)

for any vertex u and any two adjacent vertices v,w at distance k from u, there exists a common neighbor x of v,w at distance k1 from u;
 (quadrangle condition)

for any vertices u,z at distance k appart and any two neighbors v,w of z at distance k1 from u, there exists a common neighbor x of v,w at distance k2 from u.
A graph is quasimedian if it weakly modular and does not contain K_{4}^{} and K_{3,2} as induced subgraphs (see Figure 2).
In the context of CAT(0) cube complexes, combinatorially convex subcomplexes play an important role. For quasimedian graphs, this role will be played by gated subgraphs.
Definition 2.2.
Let X be a graph and Y\subset X a subgraph. Fixing a vertex x\in X, we say that a vertex p\in Y is a gate for x if, for every y\in Y, there exists a geodesic between x and y passing through p. If any vertex of X admits a gate in Y, we say that Y is gated.
It is worth noticing that:
Fact 2.3.
Let X be a graph and Y\subset X a subgraph.

•
A gate in Y of some vertex x\in X, when it exists, is the unique vertex of Y minimising the distance to x.

•
If Y is gated, then it is convex.
Proof.
Fix a vertex x\in X, and suppose that it admits a gate y\in Y. For every z\in Y, we know that
d(x,z)=d(x,y)+d(y,z)\geq d(x,y). 
Therefore, y minimises the distance to x in Y. It also follows from this inequality that, if z\in Y is another vertex minimising the distance to x, necessarily d(y,z)=0, ie., y=z. This proves the first assertion of our lemma. Now, suppose that Y is gated, and fix three vertices x,y,z\in X such that x,y are vertices of Y and z belongs to a geodesic between x,y. If p\in Y denotes the gate of z, then
d(x,y)=d(x,z)+d(z,y)=d(x,p)+2d(p,z)+d(p,y). 
On the other hand, we know from the triangle inequality that d(x,y)\leq d(x,p)+d(p,y). Consequently, d(p,z) must be zero, so that z=p\in Y. ∎
In general, it is difficult to determine whether or not a subgraph is gated just by applying the definition. Therefore, our first goal is to find a criterion which is easy to verify. We begin with a few definitions.
Definition 2.4.
Let X be a graph and Y\subset X a subgraph. If any triangle of X, ie., any three pairwise adjacent vertices of X, which shares an edge with Y is contained into Y, we say that Y contains its triangles.
Definition 2.5.
Let X be a graph and Y\subset X a subgraph. We say that Y is locally convex if any square in X with two adjacent edges contained in Y is necessarily included into Y.
The following proposition is our main criterion to prove that a subgraph is gated. The statement also follows from different results proved in [Che89]. We give a direct proof for completeness.
Proposition 2.6.
Let X be a weakly modular graph and Y\subset Y a connected subgraph. Then Y is gated if and only if it is locally convex and contains its triangles.
We begin by proving a weaker statement.
Lemma 2.7.
Let X be a weakly modular graph and Y\subset X a connected subgraph. If Y is locally convex and contains its triangles, then it is geodesic (ie., between any two vertices of Y there exists a geodesic in X lying in Y).
Proof.
Because Y is connected, it is sufficient to prove that, for every vertices x,y,z\in Y with y,z adjacent in Y such that there exists a geodesic between x and y lying in Y, necessarily there exists a geodesic between x and z lying in Y. We will argue by induction on d(x,y). Of course, for d(x,y)=0, there is nothing to prove.
Let \gamma\subset Y be a geodesic between x and y. For convenience, set \ell=d(x,y). Notice that d(x,z)d(x,y)\leq 1, so three cases may happen.

1.
Suppose that d(x,z)=\ell+1. Then, the concatenation of \gamma with the edge (y,z) between y and z defines a geodesic between x and z lying in Y.

2.
Suppose that d(x,z)=\ell. By the triangle condition, there exists a vertex a\in X adjacent to both y and z such that d(x,a)=\ell1. Notice that, because Y contains its triangles, the triangle defined by y,z,a must be included into Y. Since d(x,m)<d(x,y), we can apply our induction hypothesis to a find a geodesic between x and a which lies in Y; if we concatenate it with the edge (a,z), we find a geodesic between x and z lying in Y.

3.
Suppose that d(x,z)=\ell1. Let a be the vertex of \gamma defined by d(x,a)=\ell1. Notice first that, if a=z, then taking a subsegment of \gamma suffices to produce a geodesic between x and z lying in Y. So we will suppose that a\neq z. Then the quadrangle condition implies that there exists a vertex m\in X adjacent to both a and z such that d(x,m)=\ell2. Notice that, since Y is locally convex, the square defined by a,m,y,z must be included into Y. By applying our induction hypothesis, there exists a geodesic between x and m lying in Y, so that, by concatenating it with the edge (m,z), we find a geodesic between x and z which lies in Y.
We conclude that Y is indeed geodesic. ∎
Proof of Proposition 2.6..
It is clear that a gated subgraph must be locally convex and must contain its triangles. So we assume that our subgraph Y is locally convex and that it contains its triangles, and we want to prove that it turns out to be a gated subgraph.
Suppose by contradiction that Y is not gated. Let x\in X be a vertex which has no gate in Y. Thus, if y\in Y denotes a vertex minimizing in Y the distance to x, there exists a vertex z\in Y such that no geodesic between x and z passes through y. Without loss of generality, we may require our counterexample to minimize the quantity d(y,z), so that, for every w\in Y satisfying d(y,w)<d(y,z), there exists a geodesic between x and w passing through y. In particular, if we fix a geodesic (u_{1},\ldots,u_{r},u_{r+1}) from u_{1}=y to u_{r+1}=z, then d(x,u_{i})=d(x,y)+d(y,u_{i}) for every 1\leq i\leq r. According to the previous lemma, we can choose our geodesic between y and z in Y. Because d(x,z)d(x,u_{r})\leq d(z,u_{r})=1, three cases may happen.
Suppose that d(x,z)d(x,u_{r})=1. Then the concatenation of a geodesic between x and y with (u_{1},\ldots,u_{r+1}) would produce a geodesic between x and z passing through y, contradicting our choice of z.
Suppose that d(x,z)d(x,u_{r})=1. Because d(x,u_{r+1})=d(x,u_{r})1=d(x,u_{r1}), the quadrangle condition implies that there exists a vertex v_{r} adjacent to both u_{r1} and u_{r+1} such that d(x,v_{r})=d(x,u_{r})2. Moreover, since Y is locally convex, the square defined by u_{r},u_{r+1},v_{r},u_{r1} must be included into Y, so that v_{r}\in Y. Similarly, because d(x,v_{r})=d(x,u_{r})2=d(x,u_{r2}), the quadrangle condition implies that there exists a vertex v_{r1} adjacent to both u_{r2} and v_{r} such that d(x,v_{r1})=d(x,u_{r1})2; moreover, since Y is locally convex, the square defined by u_{r1},v_{r},v_{r1},u_{r2} must be included into Y, so that v_{r1}\in Y. Then, it is possible to iterate the argument with u_{r3} and v_{r1}, and so on. Thus, we have constructed a sequence of vertices v_{2},\ldots,v_{r}\in Y satisfying d(x,v_{i})=d(x,u_{i})2. It is worth noticing that, because y minimizes the distance to x in Y, necessarily r\geq 2, so at least v_{2} exists. But then
d(x,v_{2})=d(x,u_{2})2=d(x,u_{1})1=d(x,y)1, 
contradicting the choice of y.
Finally, suppose that d(x,z)d(x,u_{r})=0. So d(x,u_{r})=d(x,u_{r+1}), and we deduce from the triangle condition that there exists a vertex m\in X adjacent to both u_{r} and u_{r+1} such that d(x,m)=d(x,u_{r})1. Notice that, since Y contains its triangles, necessarily m\in Y. Finally, we get a contradiction by replacing u_{r+1} with m in the previous argument. ∎
We conclude this section by proving some general properties of quasimedian graphs which will be useful in the sequel. We begin by noticing that gated subgraphs satisfy the Helly’s property.
Proposition 2.8.
Let X be a graph and Y_{1},\ldots,Y_{n}\subset X a collection of gated subgraphs. If Y_{p}\cap Y_{q}\neq\emptyset for every 1\leq p,q\leq n, then \bigcap\limits_{i=1}^{n}Y_{i}\neq\emptyset.
Proof.
We will suppose that n=3. The general case follows by induction. Fix some vertices x\in Y_{1}\cap Y_{2}, y\in Y_{2}\cap Y_{3}, z\in Y_{3}\cap Y_{1}, and let m\in Y_{3} be the gate of x in Y_{3}. Because m belongs to a geodesic between x and y, we deduce from the convexity of Y_{2} that m\in Y_{2}; similarly, because m belongs to a geodesic between x and z, we deduce from the convexity of Y_{1} that m\in Y_{3}. Thus, m belongs to Y_{1}\cap Y_{2}\cap Y_{3}, so that Y_{1}\cap Y_{2}\cap Y_{3}\neq\emptyset. ∎
The following lemma follows easily from the definition of quasimedian graphs. We leave the proof as an exercice.
Lemma 2.9.
Let X be a quasimedian graph and Y\subset X a subgraph. If Y is a convex subgraph, then it is a quasimedian graph on its own right.
For our next lemma, recall that a clique is a maximal complete subgraph.
Lemma 2.10.
In a quasimedian graph, a clique contains its triangles.
Proof.
Let C be a clique, and x,y,z three pairwise adjacent vertices with x,y\in C. If C contains no other vertices than x,y and possibly z, then in fact z must belong to C since x,y,z define a complete subgraph containing C. Otherwise, suppose that there exists some vertex v\in C different from x,y,z. Then the subgraph generated by v,x,y,z contains a K_{4}^{}; since this subgraph cannot be induced by the definition of quasimedian graphs, we deduce that v and z must be adjacent. Thus, we conclude that z is ajdacent to any vertex of C, which implies that z belongs to C. ∎
In the next section, we will use this lemma to deduce that cliques are gated subgraphs (see Lemma 2.16).
Lemma 2.11.
In a quasimedian graph, the intersection between two different cliques is either empty or a single vertex.
Proof.
Let C_{1},C_{2} be two cliques intersecting along an edge e. Fix two vertices x_{1}\in C_{1}, x_{2}\in C_{2} which are not endpoints of e. The subgraph generated by e, x_{1} and x_{2} contains a K_{4}^{}; since such a subgraph cannot be induced by definition of a quasimedian graph, we deduce that x_{1} and x_{2} must be adjacent. Therefore, the vertices of C_{1} and C_{2} are pairwise adjacent, hence C_{1}=C_{2}. ∎
A direct consequence of the previous lemma is:
Corollary 2.12.
In a quasimedian graph, there exists a unique clique containing a given edge.
Essentially, our last lemma states that two cliques containing parallel edges must be parallel themselves.
Lemma 2.13.
Let C_{1},C_{2}\subset X be two distinct cliques. Suppose that there exist two edges e_{1}\subset C_{1}, e_{2}\subset C_{2} which are opposite sides of some square in X. Then the subgraph generated by C_{1}\cup C_{2} is isomorphic to C_{1}\times[0,1], where C_{1}=C_{1}\times\{0\} and C_{2}=C_{1}\times\{1\}.
Proof.
Let e_{i}=(x_{i},y_{i}) for i=1,2, where x_{1} and x_{2} (resp. y_{1} and y_{2}) are adjacent in the square defined by e_{1} and e_{2}.
First, notice that C_{1} and C_{2} are disjoint. Indeed, suppose that there exists a vertex v\in C_{1}\cap C_{2}. Then the vertices v, x_{1} and x_{2} are pairwise adjacent, hence x_{2}\in C_{1}. Similarly, we show that y_{2}\in C_{1}, so that the edge e_{2} belongs both to C_{1} and C_{2}. We deduce from Corollary 2.12 that C_{1}=C_{2}. Therefore, because we supposed C_{1} and C_{2} distinct, they are necessarily disjoint.
Then, we notice that any vertex of C_{1} is adjacent to some vertex of C_{2}. Let v\in C_{1} be a vertex, which we will suppose different from x_{1} and y_{1} since we already know that x_{1} and y_{1} are adjacent to some vertices of C_{2}. We claim that d(v,x_{2})=d(v,y_{2})=2. Indeed, if v is adjacent to x_{2} or y_{2}, say to x_{2}, then the vertices v,x_{1},x_{2} define a triangle which must be included into C_{1} since a clique contains its triangles; however, we know that C_{1} and C_{2} are disjoint, so this is impossible. Thus, the triangle condition implies that there exists a vertex m\in X adjacent to both x_{2} and y_{2} such that d(v,m)=1. Because C_{2} contains its triangles, the triangle defined by the vertices m,x_{2},y_{2} must be included into C_{2}, and in particular m\in C_{2}. We have proved that v is adjacent to some vertex of C_{2}.
On the other hand, because C_{2} contains its triangles according to Lemma 2.10 and that C_{1} and C_{2} are disjoint, a vertex of C_{1} may be adjacent to at most one vertex of C_{2}. Therefore, any vertex of C_{1} is adjacent to exactly one vertex of C_{2}. Similarly, we prove by symmetry that any vertex of C_{2} is adjacent to exacly one vertex of C_{1}. Our claim follows. ∎
2.2 Hyperplanes and sectors
In this section, we define a notion of hyperplanes in quasimedian graphs by following the definition introduced by Sageev in [Sag95] for CAT(0) cube complexes. Our goal is to prove that hyperplanes in CAT(0) cube complexes and hyperplanes in quasimedian graphs have essentially the same behaviour. A posteriori, we know that our definition coincides with the transitive closure of the \thetaequivalence introduced by Djokovic in [Djo73]; in particular, a few of the results proved in this section were also proved in [BMW94] using the \thetaequivalence.
Definition 2.14.
A hyperplane is an equivalence class of edges, where two edges e and e^{\prime} are said equivalent whenever there exists a sequence of edges e_{0}=e,e_{1},\ldots,e_{n1},e_{n}=e^{\prime} such that, for every 1\leq i\leq n1, either e_{i} and e_{i+1} are opposite sides of some square or they are two sides of some triangle.
Alternatively, if we say that two cliques are parallel whenever they respectively contain two opposite sides of some square, then a hyperplane is the collection of edges of some class of cliques with respect to the transitive closure of being parallel.
One says that an edge or a clique is dual to a given hyperplane if it belongs to the associated class of edges. Of course, because two distinct equivalence classes are necessarily disjoint, an edge or a clique is dual to a unique hyperplane.
From Lemma 2.13, we know that two cliques C_{1},C_{2} are parallel if and only if there exists an induced subgraph C\times[0,1] where C\times\{0\}=C_{1} and C\times\{1\}=C_{2}.
We sum up the results of this section in the next proposition (see also Figure 3); the needed definitions are given progressively below. The third point is contained in Corollary 2.21 and Corollary 2.22; the first point is contained in Lemma 2.24 and Lemma 2.28; and finally the second point is contained in Lemma 2.29. We also use these results to characterize geodesics, see Proposition 2.30.
Proposition 2.15.
Let X be a quasimedian graph and J a hyperplane.

•
The carrier N(J) of J is a gated subgraph isomorphic to F(J)\times C, where F(J) is the main fiber of J and C is a clique.

•
Every connected component of \partial J, called a fiber, is a gated subgraph isomorphic to F(J); in particular, F(J) is a quasimedian graph on its own right.

•
The hyperplane J separates X into at least two connected components, called sectors. They are gated subgraphs.
First, we want to define what is a sector.
Lemma 2.16.
In a quasimedian graph, any clique is gated.
Proof.
Let X be a quasimedian graph and C\subset X a clique. According to Proposition 2.6 and Lemma 2.10, it is sufficient to prove that C is locally convex in order to deduce that C is gated.
Let \Pi be a square in X with two adjacent edges in C. Because C is complete, this implies that \Pi admits a diagonal in X, defining a subgraph isomorphic to K_{4}^{}; then, since such a subgraph cannot be induced, the second diagonal of \Pi must exist in X, so the subgraph generated by \Pi is complete. Because we already know that C contains its triangles, we deduce that \Pi\subset C. Therefore, C is locally convex. ∎
Definition 2.17.
Let X be a quasimedian graph, C\subset X a clique and v\in C a vertex. The subgraph [C,v] generated by the set of all the vertices of X whose gate in C is v is a sector. We refer to the collection \{[C,v]\mid v\in C\} as the sector decomposition defined by C.
Lemma 2.18.
Let C_{1},C_{2}\subset X be two cliques. If C_{1} and C_{2} are dual to the same hyperplane, their sector decompositions are the same.
Proof.
If C_{1}=C_{2} there is nothing to prove. Otherwise, there exists a sequence of cliques between C_{1} and C_{2} such that any two successive cliques are parallel. We will assume that C_{1} and C_{2} are parallel, the general case following by induction.
We want to prove the following statement, which is sufficient to conclude. For every vertex x\in C_{1}, if p denotes the vertex of C_{2} opposite to x, then [C_{1},x]=[C_{2},p]. In fact, we only need to prove the inclusion [C_{1},x]\subset[C_{2},p], the reverse inclusion following by symmetry. More precisely, for any a\in[C_{1},x] and q\in C_{2}\backslash\{p\}, we claim that d(a,p)<d(a,q), meaning that p is the gate of a in C_{2}.
Let y\in C_{1} denote the vertex opposite to q. Notice that neither x and q nor y and p are adjacent, since a clique must contain its triangles and that C_{1} and C_{2} are disjoint. For convenience, let k=d(a,x). Because d(a,p)d(a,x)\leq d(p,x)=1, we know that d(a,p)\in\{k1,k,k+1\}. Notice moreover that
d(a,x)+1=d(a,x)+d(x,y)=d(a,y)\leq d(a,q)+d(q,y)=d(a,q)+1 
implies d(a,q)\geq d(a,x)=k. Therefore, if we suppose that d(a,q)\leq d(a,p), only three cases may happen.
Suppose that d(a,p)=k and d(a,q)=k. Because d(a,x)=k=d(a,q) and d(a,y)=k+1, the quadrangle condition implies that there exists a vertex m\in X adjacent to both x and q such that d(a,m)=k1. In particular, d(a,m)=k1 implies that m\notin\{p,q\}. Notice that the vertices x,y,p,q,m define a subgraph isomorphic to K_{2,3}, so it cannot be induced. However, we know that x and q are not adjacent, and
k+1=d(a,y)\leq d(a,m)+d(m,y)=k1+d(m,y) 
implies d(m,y)\geq 2, ie., m and y are not adjacent. Therefore, either y and p or m and p must be adjacent. If y and p are adjacent, then the vertices p,q,x,y define an induced subgraph isomorphic to K_{4}^{}; similarly, if m and p are adjacent, then the vertices p,q,x,m define an induced subgraph isomorphic to K_{4}^{}. Thus, we get a contradiction.
Suppose that d(a,p)=k+1 and d(a,q)=k. Because d(a,x)=k=d(a,q) and d(a,y)=k+1, the quadrangle condition implies that there exists a vertex m\in X adjacent to both x and q such that d(a,m)=k1. Then exactly the same argument as above produces a contradiction.
Suppose that d(a,p)=k+1 and d(a,q)=k+1. By applying the triangle condition, we find a vertex r adjacent to both p and q such that d(a,r)=k. Since the clique C_{2} contains its triangles, necessarily r\in C_{2}. Let s denote the vertex of C_{1} opposite to r. A fortiori, d(a,s)=d(a,x)+d(x,s)=k+1. Thus, by replacing y and q with s and r respectively, we produce the siutation that occurs in our first case. In particular, we deduce a contradiction as well.
As a consequence, we conclude that necessarily d(a,q)>d(a,p). ∎
Definition 2.19.
Let X be a quasimedian graph and J a hyperplane. If C is a clique dual to J, the sectors delimited by J are the sectors delimited by C.
According to Lemma 2.18, this definition does not depend on the choice of the clique C.
Lemma 2.20.
Let X be a quasimedian graph, J a hyperplane and e\subset X an edge. The endpoints of e belong to the same sector delimited by J if and only if e\notin J.
Proof.
Let C be a clique dual to J, x,y\in X the endpoints of e, and a,b\in C their gates in C respectively. Suppose that x and y does not belong to the same sector, so that a and b are two distinct adjacent vertices. Notice that
1+d(x,a)=d(x,b)\leq d(x,y)+d(y,b)=1+d(y,b), 
hence d(x,a)\leq d(y,b). Similarly, we show that d(y,b)\leq d(x,a), hence d(x,a)=d(y,b); let \ell denote this common value. Let x_{0},x_{1},\ldots,x_{m} be a geodesic between x_{0}=x and x_{m}=a, and set y_{0}=y. Notice that x_{1}\neq y because x_{1} and y have different gates in C, d(b,y)=\ell, d(b,x)=\ell+1 and
d(b,x_{1})=d(b,a)+d(a,x_{1})=1+\ell1=\ell, 
so the quadrangle condition implies that there exists a vertex y_{1}\in X adjacent to both x_{1} and y such that d(b,y_{1})=d(b,x)2=d(b,y)1. Morever, the gate of y_{1} in C is again b. Indeed,
1+d(y_{1},b)=d(y,b)=d(y,C)\leq 1+d(y_{1},C), 
so b minimizes in C the distance to y_{1}. Then, we can define y_{2} similarly from x_{2} and y_{1}, and so on. Thus, we have constructed a sequence y_{0},y_{1},\ldots,y_{m}\in X such that the gate of y_{i} in C is b, and x_{i},y_{i},y_{i+1},x_{i+1} define a square, and d(b,y_{i})=d(b,y)i. In particular, (x_{m},y_{m})=(a,b) so we conclude that the edge e=(x,y) is dual to J.
Conversely, suppose that e\in J. Let C^{\prime} denote the unique clique containing the edge (x,y). Clearly, x and y belong to different sectors delimited by C^{\prime}, so, because J is dual to C^{\prime}, they have to belong to different sectors delimited by J. ∎
Corollary 2.21.
Let X be a quasimedian graph and J a hyperplane. Let X\backslash J denote the graph obtained from X by removing the interiors of the edges of J. The connected components of X\backslash J are precisely the sectors delimited by J.
Proof.
Let C be a clique dual to J. According to Lemma 2.20, we know that any edge of X either belongs to a sector delimited by C or is dual to J, so we only have to show that two sectors delimited by C are separated by J and that a sector is connected.
Let u,v\in C be two distinct vertices. If \gamma is a path between two vertices of [C,u] and [C,v], then there necessarily exists an edge e\subset\gamma whose endpoints have different gates in C. It follows from Lemma 2.20 that e is dual to J, hence \gamma\nsubseteq X\backslash J. Therefore, two sectors delimited by C are separated by J.
The connectedness of a sector follows from the observation that any vertex can be joined by a geodesic to its gate in C, and any vertex of this geodesic has necessarily the same gate in C. Such a path does not contain any edge of J according to Lemma 2.20. ∎
Corollary 2.22.
In a quasimedian graph, a sector is gated.
Proof.
Let C be a clique and v\in C a vertex, and let J denote the hyperplane dual to C. According to Proposition 2.6, it is sufficient to prove that [C,v] is locally convex and contains its triangles in order to deduce that it is gated.
Let (a,b,c) be a triangle with (a,b)\subset[C,v]. Because a and b belong to the same sector delimited by J, it follows from Lemma 2.20 that (a,b)\notin J. Necessarily, (a,c)\notin J. Lemma 2.20 implies that a and c belong to the same sector delimited by J, hence c\in[C,v], and finally (a,b,c)\subset[C,v].
Let (a,b,c,d) be a square with (a,b),(b,c)\subset[C,v]. Because a and b belong to the same sector delimited by J, it follows from Lemma 2.20 that (a,b)\notin J. A fortiori, (c,d)\notin J. Therefore, Lemma 2.20 implies that c and d belong to the same sector delimited by J, hence d\in[C,v], and finally (a,b,c,d)\subset[C,v]. ∎
Definition 2.23.
The carrier of a hyperplane J of X, denoted by N(J), is the subgraph generated by the endpoints of the edges which belong to J. The boundary of J, denoted by \partial J, is the graph obtained from N(J) by removing the interiors of the edges which belong to J.
Lemma 2.24.
In a quasimedian graph, the carrier of a hyperplane is gated.
We begin by proving two preliminary lemmas. Using the vocabulary of CAT(0) cube complexes, the first lemma below states that hyperplanes do not selfintersect nor selfosculate.
Lemma 2.25.
Let C_{1},C_{2}\subset X be two distinct cliques dual to the same hyperplane. Either C_{1}=C_{2} or C_{1}\cap C_{2}=\emptyset.
Proof.
Suppose that there exists some vertex u\in C_{1}\cap C_{2}. Let v\in C_{1}\backslash\{u\} be a vertex; or course, u and v are adjacent. If d(v,C_{2})=1, then u must be the gate of v in C_{2}, so that u and v belong to the same sector delimited by C_{2}, contradicting Lemma 2.20 because the edge (u,v) is dual to the same hyperplane as the clique C_{2}. Therefore, d(v,C_{2})=0. This implies that the edge (u,v) is included into C_{1}\cap C_{2}, hence C_{1}=C_{2} as a consequence of Lemma 2.11. ∎
Lemma 2.26.
Let C_{1},C_{2}\subset X be two distinct cliques dual to the same hyperplane. If there exists an edge between C_{1} and C_{2}, then C_{1} and C_{2} are parallel.
Proof.
Let u\in C_{1} and v\in C_{2} be two adjacent vertices and let w\in C_{1}\backslash\{u\}. Notice that, as a consequence of Lemma 2.25, C_{1} and C_{2} must be disjoint. First, we claim that d(v,w)=2. Indeed, because C_{1}\cap C_{2}=\emptyset, necessarily d(v,w)\geq 1; moreover, because a clique contains its triangles, d(v,w)=1 would imply v\in C_{1}\cap C_{2}. Thus, we have d(v,w)=2, and in particular d(w,C_{2})\leq 2. On the other hand, once again because C_{1} and C_{2} are disjoint, necessarily d(w,C_{2})\geq 1; and d(w,C_{2})=2 would imply that v is the gate of u and w in C_{2}, so that u and w would belong to the same sector delimited by C_{2}, contradiction Lemma 2.20. Therefore, d(w,C_{2})=1, ie., there exists some vertex x\in C_{2} adjacent to w. This proves that C_{1} and C_{2} are parallel according to Lemma 2.13. ∎
Proof of Lemma 2.24..
According to Proposition 2.6, it is sufficient to prove that N(J) is locally convex and contains its triangles in order to deduce that it is gated.
Let (a,b,c) be a triangle with (a,b)\subset N(J). If (a,b)\in J then (a,c)\in J, which implies c\in N(J). So we will suppose that (a,b)\notin J. Thus, there exist two edges (a,e),(b,f)\in J. Notice that, because (a,b)\notin J, necessarily a and f, and similarly b and e are not adjacent. A fortiori, (a,e) and (b,f) belong to distinct cliques of J, and Lemma 2.26 implies that e and f are adjacent. Once again because (a,b)\notin J, we know that c and e, and similarly c and f, are not adjacent. Therefore, d(c,e)=d(c,f)=2. The triangle condition implies that there exists a vertex m\in X adjacent to e,f,c. Clearly, (c,m)\in J hence c\in N(J), and finally (a,b,c)\subset N(J).
Let (a,b,c,d) be a square with (a,b),(b,c)\subset N(J). If (a,b)\in J then (d,c)\in J, which implies d\in N(J); similarly, if (b,c)\in J then (a,d)\in J, which implies d\in N(J). So we suppose that (a,b),(b,c)\notin J. In particular, we know that there exist three edges (a,e),(b,f),(c,g)\in J. Notice that, because (a,b)\notin J, necessarily a is not adjacent to f and b is not adjacent to e. Therefore, (a,e) and (b,f) belong to distinct cliques of J so that Lemma 2.26 implies that e and f are adjacent. Similarly, we show that f and g are adjacent. Now, notice that (b,c)\notin J implies that d and e are not adjacent, hence d(e,d)=2; similarly, we show that d(g,d)=2. Let J_{1} denote the hyperplane dual to (a,b) and J_{2} the hyperplane dual to (b,c). Because (a,b)\notin J and (b,c)\notin J, we know that J_{1}\neq J and J_{2}\neq J. If J_{1}=J_{2}, then this hyperplane is dual to the two edges (c,d) and (c,b), and we deduce from Lemma 2.25 that these edges must belong to the same clique, so that b and d must be adjacent. On the other hand, we already know that N(J) contains its triangles, so (a,b)\subset N(J) implies d\in N(J). So suppose that J_{1}\neq J_{2}. It is worth noticing that d and f are separated by at least three (distinct) hyperplanes, namely J,J_{1},J_{2}. Indeed, it follows from Lemma 2.20 that d and b belong to the same sector delimited by J and that b and d belong to different sectors delimited by J, so that J must separate f and d; similarly, J_{1} and J_{2} separate d and f. As a consequence, necessarily d(f,d)\geq 3, and in fact d(f,d)=3 since there exists a path of length three between d and f by construction. Now we are able to deduce from the quadrangle condition that there exists a vertex m\in X adjacent to d,e,g. Clearly, the edge (d,m) belongs to J, hence d\in N(J). Consequently, (a,b,c,d)\subset N(J). ∎
Definition 2.27.
The main fiber of a hyperplane J, denoted by F(J), is the graph whose vertices are the cliques of J and whose edges link two parallel cliques.
Since any vertex of N(J) belongs to a unique clique of J, as a consequence of Lemma 2.25, there exists a canonical projection c:N(J)\to F(J). Furthermore, if we fix some clique C dual to J and denote by p:X\to C the application which associates to a vertex of X its gate in C, we can define a natural map
\Psi:\left\{\begin{array}[]{ccc}N(J)&\longrightarrow&F(J)\times C\\ v&\longmapsto&(c(v),p(v))\end{array}\right. 
Lemma 2.28.
\Psi defines an isomorphism between the graphs N(J) and F(J)\times C.
Proof.
First, we prove that \Psi sends adjacent vertices to adjacent vertices. Let e=(x,y)\subset N(J) be an edge. If e\in J then x and y belong to the same clique of J, ie., c(x)=c(y); moreover, Lemma 2.20 implies that p(x)\neq p(y), so that p(x) and p(y) are necessarily adjacent. A fortiori, \Psi(x) and \Psi(y) are adjacent. If e\notin J, Lemma 2.20 implies that p(x)=p(y). Moreover, x and y necessarily belong to distinct cliques of J, which are linked by an edge since x and y are adjacent; so it follows from Lemma 2.26 that these cliques are parallel, ie., c(x) and c(y) are adjacent in F(J). A fortiori, \Psi(x) and \Psi(y) are adjacent.
We claim that \Psi is injective. Let u,v\in N(J) be two distinct vertices. If c(u)\neq c(v), necessarily \Psi(u)\neq\Psi(v), so we suppose that c(u)=c(v), ie., u and v belong to the same clique of J. In particular, u and v are adjacent, and the edge (u,v) belongs to J. It follows from Lemma 2.20 that p(u)\neq p(v). A fortiori, \Psi(u)\neq\Psi(v).
We claim that \Psi is surjective. Let (C^{\prime},x)\in F(J)\times C. According to Lemma 2.18, there exists a vertex u\in C^{\prime} such that [C^{\prime},u]=[C,x]. In particular, we know that u\in[C,x], hence p(u)=x. Finally, \Psi(u)=(C^{\prime},x).
In order to conclude that \Psi is an isomorphism, it remains to verify that, if u,v\in N(J) are vertices, then \Psi(u) adjacent to \Psi(v) implies that u is adjacent to v. So suppose that \Psi(u) and \Psi(v) are adjacent. Two cases may happen. First, suppose that c(u)=c(v) and that p(u) is adjacent to p(v). Then u and v belong to the same clique of J, and p(u)\neq p(v) implies u\neq v, so that u is adjacent to v. Now, suppose that p(u)=p(v) and that c(u) is adjacent to c(v). Thus, u and v belong to the same sector delimited by J, and to two parallel cliques of J, say u\in C_{1} and v\in C_{2}. Let x denote the vertex of C_{1} opposite to v. According to Lemma 2.25, the edge (x,v) does not belong to J, so that x and v belong to the same sector delimited by J according to Lemma 2.20. A fortiori, u and x necessarily belong to the same sector delimited by J. Because u and x belong to the same clique of J, we deduce from Lemma 2.20 that u=x. As a consequence, u and v are adjacent. ∎
Lemma 2.29.
Let F be a connected component of \partial J. Then F is a gated subgraph isomorphic to the main fiber F(J). More precisely, there exists a vertex v\in C such that F=\Psi^{1}(F(J)\times\{v\}).
Proof.
Noticing that \Psi(J)=\bigcup\limits_{K\in F(J)}\{K\}\times C, we deduce that \Psi(\partial J)=\bigcup\limits_{v\in C}F(J)\times\{v\}. As a consequence, there exists a vertex v\in C such that F=\Psi^{1}(F(J)\times\{v\}). In particular, F is clearly isomorphic to F(J). Moreover, it is worth noticing that the vertices of F are precisely the vertices of N(J) whose image by p is v, ie., F=N(J)\cap[C,v]. Since the intersection of two gated subgraphs is gated, we deduce from Corollary 2.22 and Lemma 2.24 that F is a gated subgraph. ∎
All the results above imply Proposition 2.15, as already mentionned. Now, we want to characterize geodesics from the hyperplanes they intersect. The analogue result for CAT(0) cube complexes is fundamental.
Proposition 2.30.
Let X be a quasimedian graph. A path in X is a geodesic if and only if it intersects any hyperplane at most once. In particular, the distance between two vertices of X is equal to the number of hyperplanes separating them.
Proof.
Let p be a path which intersects a hyperplane J twice. Let u be the first vertex of p which belongs to N(J), and similarly v the last vertex of p which belongs to N(J). Let q denote the subpath of p between u and v. Since N(J) is convex, according to Lemma 2.24, if q\nsubseteq N(J) then p cannot be a geodesic. So suppose that q\subset N(J). Let \Psi:N(J)\to F(J)\times C denote the isomorphism given by Lemma 2.28. Clearly, \Psi(q) is not a geodesic in F(J)\times C since q contains two edges dual to J, so q=\Psi(\Psi^{1}(q)) cannot be geodesic in X. A fortiori, p cannot be a geodesic as well. Thus, we have proved that a geodesic must intersect any hyperplane at most once.
Let x,y\in X be two vertices and \gamma a geodesic between them. If J is a hyperplane which does not separate x and y, it follows from the convexity of the sectors given by Corollary 2.22 that \gamma must be disjoint from J. Therefore, any hyperplane intersecting \gamma necessarily separates x and y. On the other hand, we know that \gamma intersects at most once each of these hyperplanes, so its length is bounded above by the number of hyperplanes separating x and y. Conversely, \gamma necessarily intersects any hyperplane separating x and y, so its length is bounded below by the number of hyperplanes separating x and y. Therefore, the distance between x and y is precisely the number of hyperplanes separating them.
Now, let p be a path between two vertices x,y\in X such that p intersects any hyperplane at most once. If J is a hyperplane which does not separate x and y, necessarily p must be disjoint from J since otherwise it would have to intersect it at least twice because x and y lie in different sectors delimited by J. Therefore, p intersects only the hyperplanes separating x and y. Because p intersects any hyperplane at most once, we deduce that its length is bounded above by the number of hyperplanes separating x and y, which correponds exactly to the distance between x and y. Therefore, p has to be a geodesic. ∎
2.3 Projections onto gated subgraphs
In this section, we generalize the results proved in [Gen16c, Section 2] on combinatorial projections in CAT(0) cube complexes.
Definition 2.32.
Let X be a quasimedian graph and Y\subset X a gated subgraph. The application \mathrm{proj}_{Y}:X\to Y which sends a vertex of X to its gate in Y will be referred to as the projection onto Y.
Our first main result describes the hyperplanes separating the projections of two vertices.
Proposition 2.33.
Let Y\subset X be a gated subgraph and let p:X\to Y denote the projection onto Y. For every vertices x,y\in X, the hyperplanes separating p(x) and p(y) are precisely the hyperplanes which separate x and y and intersect Y.
Our proposition will be a consequence of the following lemma:
Lemma 2.34.
Let Y\subset X be a gated subgraph and x\in X a vertex. Any hyperplane separating x and its projection onto Y separates x and Y.
Proof.
Suppose by contradiction that there exists a hyperplane separating x and its projection p onto Y which intersects Y. In particular, J must separate p from some vertex y\in Y. As a consequence, if we fix a geodesic \gamma between x and y passing through p, necessarily \gamma has to intersect \gamma at least twice. This contradicts Proposition 2.30. ∎
Proof of Proposition 2.33..
Let J be a hyperplane separating p(x) and p(y). In particular, J intersects Y. Because the previous lemma implies that any hyperplane separating x and p(x) or y and p(y) must be disjoint from Y, we deduce that J separates neither x and p(x) nor y and p(y). Thus, J must separate x and y. Conversely, suppose that J is a hyperplane which separates x and y and intersects Y. The same argument shows that J cannot separate x and p(x) or y and p(y), so that J must separate p(x) and p(y). ∎
Corollary 2.35.
Let Y\subset X be a gated subgraph and let p:X\to Y denote the projection onto Y. Then p is 1Lipschitz, ie., d(p(x),p(y))\leq d(x,y) for every vertices x,y\in X.
Proof.
Although Lemma 2.34 has been introduced to prove Proposition 2.33, it will turn out to be quite useful on its own right. For instance, it leads to the following result:
Lemma 2.36.
Let Y_{1},Y_{2}\subset X be two gated subgraphs. If x\in Y_{1} and y\in Y_{2} are two vertices minimizing the distance between Y_{1} and Y_{2}, then the hyperplanes separating x and y are precisely the hyperplanes separating Y_{1} and Y_{2}.
Proof.
Let p:X\to Y_{1} and q:X\to Y_{2} denote respectively the projections onto Y_{1} and Y_{2}. In particular, we have y=p(x) and x=q(y). By applying Lemma 2.34 twice, we deduce that any hyperplane separating x and y must be disjoint from Y_{1} and Y_{2}; a fortiori, it has to separate Y_{1} and Y_{2}. Conversely, it is clear that any hyperplane separating Y_{1} and Y_{2} must separate x and y. ∎
Our second main result is the following:
Proposition 2.37.
Let Y_{1},Y_{2}\subset X be two gated subgraphs, and let p:X\to Y_{2} denote the projection onto Y_{2}. Then p(Y_{1}) is a geodesic subgraph of Y_{2}, and the hyperplanes intersecting p(Y_{1}) are precisely the hyperplanes intersecting both Y_{1} and Y_{2}.
Proof.
According to Corollary 2.35, p is 1Lipschitz, so p sends an edge to either an edge or a vertex; in particular, p sends a path to a path. Fix two vertices x,y\in X and a geodesic \gamma between them; let \bar{\gamma}\subset p(Y_{1}) be the path which is the image of \gamma by p. According to Proposition 2.30, it is sufficient to prove that no hyperplane intersects twice \bar{\gamma} in order to justify that \bar{\gamma} is a geodesic. If e_{1} and e_{2} are two edges of \bar{\gamma}, choose two lifts f_{1},f_{2}\subset\gamma. It follows from Proposition 2.33 that the hyperplane dual to e_{1} (resp. e_{2}) is also dual to f_{1} (resp. f_{2}). Therefore, a hyperplane dual to both e_{1} and e_{2} would be dual to both f_{1} and f_{2}, and so would intersect \gamma twice, which contradicts the fact that \gamma is a geodesic. Thus, we have proved that p(Y_{1}) is a geodesic subgraph.
If a hyperplane intersects p(Y_{1})\subset Y_{2}, it cleary intersects Y_{2} and it must intersect Y_{1} according to Proposition 2.33. Conversely, if a hyperplane J intersects both Y_{1} and Y_{2}, it is dual to an edge e\subset Y_{1}, and it follows from Proposition 2.33 that J separates the projections onto Y_{2} of the endpoints of e; a fortiori, J intersects p(Y_{1}). ∎
We conclude this section by proving the following result which will be useful later.
Lemma 2.38.
Let X be a quasimedian graph and Y_{1},Y_{2}\subset X two intersecting gated subgraphs. Then \mathrm{proj}_{Y_{2}}\circ\mathrm{proj}_{Y_{1}}=\mathrm{proj}_{Y_{1}\cap Y_{2}}.
We begin by proving the following observation.
Lemma 2.39.
Let X be a quasimedian graph, Y_{1},Y_{2}\subset X two intersecting gated subgraphs, and y\in Y_{1} a vertex. If z denotes the projection of y onto Y_{2}, then z\in Y_{1}\cap Y_{2}.
Proof.
If we fix some vertex w\in Y_{1}\cap Y_{2}, then z must belong to a geodesic between y and w, hence z\in I(y,w). On the other hand, because Y_{1} is convex, necessarily I(y,w)\subset Y_{1}, hence z\in Y_{1}. We already know that z\in Y_{2}, so z\in Y_{1}\cap Y_{2}. ∎
Proof of Lemma 2.38..
Let x\in X be a vertex. Let y denote its projection onto Y_{1} and z the projection of y onto Y_{2}. We want to prove that z is the projection of x onto Y_{1}\cap Y_{2}. Notice that z belongs to Y_{1}\cap Y_{2} according to the previous lemma.
Now, let w\in Y_{1}\cap Y_{2} be any vertex. Noticing that
\begin{array}[]{lcl}d(x,w)&=&d(x,y)+d(y,w)\\ &=&d(x,y)+d(y,z)+d(z,w)\\ &\geq&d(x,y)+d(y,z)=d(x,z),\end{array} 
we deduce that z is the vertex of Y_{1}\cap Y_{2} minimizing the distance to x, ie., z is the projection of x onto Y_{1}\cap Y_{2}. ∎
As an immediate corollary, we deduce the following statement.
Corollary 2.40.
Let X be a quasimedian graph and Y_{1}\subset Y_{2} two gated subgraphs. Then \mathrm{proj}_{Y_{1}}\circ\mathrm{proj}_{Y_{2}}=\mathrm{proj}_{Y_{1}}.
2.4 Quasicubulating popsets
Based on works of Dunwoody and Sageev [Sag95], Roller describes in [Rol98] how to construct CAT(0) cube complexes from some special class of posets, called pocsets; see also [Sag14]. We generalize the construction to obtain quasimedian graphs. It is worth noticing that, in this paper, quasicubulating popsets will be used only to prove Proposition 2.63, from which we will prove Corollary 2.65, Lemma 2.74 and Lemma 2.110 below; and later to show how to cubulate a space with walls containing duplicates (which can also be done using pocsets). Therefore, the technicalities below can be skipped in a first lecture.
Before giving the definition of a popset, let us mention an important example to keep in mind. Let X be a quasimedian graph, and let \mathcal{S}(X) denote the set of the sectors of X. For every hyperplane J of X, let \mathcal{P}(J) denote the set of sectors delimited by J. Notice that \mathfrak{P}=\{\mathcal{P}(J)\mid J\ \text{hyperplane}\} defines a partition of \mathcal{S}(X). The data (\mathcal{S}(X),\subset,\mathfrak{P}) is an example of what we call a popset.
Definition 2.41.
A popset (X,<,\mathfrak{P}) is the data of a poset (X,<) with a partition \mathfrak{P} of X such that:

•
for every \mathcal{P}\in\mathfrak{P}, \#\mathcal{P}\geq 2;

•
for every \mathcal{P}\in\mathfrak{P}, no two elements of \mathcal{P} are <comparable;

•
for every \mathcal{P}_{1},\mathcal{P}_{2}\in\mathfrak{P}, if there exist A_{1}\in\mathcal{P}_{1} and A_{2}\in\mathcal{P}_{2} such that A_{1}<A_{2}, then there exists some B_{1}\in\mathcal{P}_{1} such that A<A_{2} for every A\in\mathcal{P}_{1}\backslash\{B_{1}\} and B<B_{1} for every B\in\mathcal{P}_{2}\backslash\{A_{2}\}. If so, we say that \mathcal{P}_{1} and \mathcal{P}_{2} are nested. (Notice that being nested is a symmetric relation.)
An element of \mathfrak{P} is referred to as a wall, and an element of a wall as a sector; notice that a sector is an element of X. Two walls which are not nested are transverse. If \mathcal{P}_{1},\mathcal{P}_{2}\in\mathfrak{P} are two nested walls and A_{1}\in\mathcal{P}_{1} is a sector, we say that A_{1} is a sector delimited by \mathcal{P}_{1} which contains \mathcal{P}_{2} if there exists some A_{2}\in\mathcal{P}_{2} such that D<A_{1} for every D\in\mathcal{P}_{2}\backslash\{A_{2}\}.
First, let us notice that there exists a unique sector of \mathcal{P}_{1} which contains \mathcal{P}_{2} when \mathcal{P}_{1} and \mathcal{P}_{2} are nested.
Lemma 2.42.
Let \mathcal{P}_{1},\mathcal{P}_{2}\in\mathfrak{P} be two nested walls and A_{1} (resp. A_{2}) a sector delimited by \mathcal{P}_{1} (resp. \mathcal{P}_{2}) which contains \mathcal{P}_{2} (resp. \mathcal{P}_{1}). Then A_{1} and A_{2} are not <comparable.
Proof.
Suppose by contradiction that A_{1}<A_{2}. Then D<A_{2} for every D\in\mathcal{P}_{1}, and there exists a sector B_{1}\in\mathcal{P}_{1} such that A<B_{1} for every A\in\mathcal{P}_{2}\backslash\{A_{2}\}. Therefore, if D\in\mathcal{P}_{2}\backslash\{A_{2}\}, then D<B_{1}<A_{2}, which is impossible since two sectors of \mathcal{P}_{2} are not <comparable. ∎
Corollary 2.43.
Let \mathcal{P}_{1},\mathcal{P}_{2}\in\mathcal{P} be two nested walls. There exists a unique sector delimited by \mathcal{P}_{1} which contains \mathcal{P}_{2}.
Proof.
Let A_{1} (resp. A_{2}) be a sector delimited by \mathcal{P}_{1} (resp. \mathcal{P}_{2}) which contains \mathcal{P}_{2} (resp. \mathcal{P}_{1}). If A\in\mathcal{P}_{1}\backslash\{A_{1}\}, then A<A_{2}. We deduce from the previous lemma that A cannot be a sector delimited by \mathcal{P}_{1} which contains \mathcal{P}_{2}. ∎
An orientation \sigma, as defined below, may be thought of as “choice function”: given a wall \mathcal{P}\in\mathfrak{P}, \sigma chooses the sector \sigma(\mathcal{P}) delimited by \mathcal{P}. Our definition mimic the definition of ultrafilters introduced in [Rol98].
Definition 2.44.
Let (X,<,\mathfrak{P}) be a popset. An orientation is a map \sigma:\mathfrak{P}\to X such that

•
\sigma(\mathcal{P})\in\mathcal{P} for every \mathcal{P}\in\mathfrak{P};

•
if A_{1}\in\mathcal{P}_{1} and A_{2}\in\mathcal{P}_{2} satisfy A_{1}<A_{2}, then A_{1}=\sigma(\mathcal{P}_{1}) implies A_{2}=\sigma(\mathcal{P}_{2}).
Definition 2.45.
Given a popset (X,<,\mathfrak{P}), we define its quasicubulation C(X,<,\mathfrak{P}) as the graph whose vertices are the orientations of (X,<,\mathfrak{P}) and whose edges link two orientations which differ only on a single wall.
Our purpose is first to show that this construction yields quasimedian graphs, and next to understand the combinatorics of its hyperplanes (see Theorem 2.56). The first step, which is Proposition 2.46 below, needs several preliminary lemmas.
Proposition 2.46.
Let (X,<,\mathfrak{P}) be a popset. Any connected component of its quasicubulation C(X,<,\mathfrak{P}) is a quasimedian graph.
For convenience, we introduce some notation. If \sigma is an orientation and D a sector of some wall \mathcal{P}\in\mathfrak{P}, we define the map
[\sigma,D]:\left\{\begin{array}[]{ccc}\mathfrak{P}&\longrightarrow&X\\ \mathcal{Q}&\longmapsto&\left\{\begin{array}[]{cl}\sigma(\mathcal{Q})&\text{if% }\ \mathcal{Q}\neq\mathcal{P}\\ D&\text{otherwise}\end{array}\right.\end{array}\right.. 
In particular, two orientations \sigma_{1} and \sigma_{2} are adjacent in C(X,\mathfrak{P}) if and only if \sigma_{2}=[\sigma_{1},D] for some sector D\neq\sigma_{1}(\mathcal{P}) (if so, notice that we also have \sigma_{1}=[\sigma_{2},\sigma_{1}(\mathcal{P})] where \mathcal{P} denotes the wall delimiting D). For convenience, we will write [\sigma,D_{1},D_{2}] instead of [[\sigma,D_{1}],D_{2}], [\sigma,D_{1},D_{2},D_{3}] instead of [[\sigma,D_{1},D_{2}],D_{3}], and so on.
Thought of as choice functions, a crucial property satisfied by any orientation is that, if it is possible to modify the choice on some wall to get an orientation, then choosing any sector will produce an orientation. This statement is made precise by Corollary 2.48.
Lemma 2.47.
Let \sigma be an orientation and A\neq\sigma(\mathcal{P}) a sector delimited by some wall \mathcal{P}. Then [\sigma,A] defines an orientation if and only if \sigma(\mathcal{P}) is minimal in \sigma(\mathfrak{P}).
Proof.
Suppose that \sigma(\mathcal{P}) is minimal in \sigma(\mathfrak{P}). To prove that [\sigma,A] defines an orientation, it is sufficient to show that, if A_{1},A_{2} are two sectors, respectively delimited by the walls \mathcal{P}_{1},\mathcal{P}_{2}, satisfying A_{1}<A_{2} and A_{1}=[\sigma,A](\mathcal{P}_{1}), then A_{2}=[\sigma,A](\mathcal{P}_{2}). Notice that \mathcal{P}_{1} and \mathcal{P}_{2} must be nested, and that A_{2} is the sector delimited by \mathcal{P}_{2} which contains \mathcal{P}_{1}; let B_{1} denote the sector delimited by \mathcal{P}_{1} which contains \mathcal{P}_{2}. Three cases may happen.
Suppose that \mathcal{P}=\mathcal{P}_{1}. If \sigma(\mathcal{P}_{1})\neq B_{1}, then \sigma(\mathcal{P}_{1})<A_{2} which implies that A_{2}=\sigma(\mathcal{P}_{2}) because \sigma is an orientation. If \sigma(\mathcal{P}_{1})=B_{1}, notice that, for every D\in\mathcal{P}_{2}\backslash\{A_{2}\}, we have D<B_{1}=\sigma(\mathcal{P}_{1})=\sigma(\mathcal{P}), so that, since \sigma(\mathcal{P}) is minimal in \sigma(\mathfrak{P}) by assumption, we deduce that D\neq\sigma(\mathcal{P}_{2}). A fortiori, \sigma(\mathcal{P}_{2})=A_{2}. We conclude that A_{2}=\sigma(\mathcal{P}_{2})=[\sigma,A](\mathcal{P}_{2}) since \mathcal{P}_{2}\neq\mathcal{P}_{1}=\mathcal{P}.
Suppose that \mathcal{P}=\mathcal{P}_{2}. We have A_{1}=[\sigma,A](\mathcal{P}_{1})=\sigma(\mathcal{P}_{1}) since \mathcal{P}_{1}\neq\mathcal{P}_{2}=\mathcal{P}. Therefore, because \sigma is an orientation, A_{1}<A_{2} implies A_{2}=\sigma(\mathcal{P}_{2}). We deduce that \sigma(\mathcal{P}_{1})=A_{1}<A_{2}=\sigma(\mathcal{P}_{2})=\sigma(\mathcal{P}), which is impossible since \sigma(\mathcal{P}) is minimal in \sigma(\mathfrak{P}) by assumption.
Suppose that \mathcal{P}\neq\mathcal{P}_{1},\mathcal{P}_{2}. Then A_{1}=[\sigma,A](\mathcal{P}_{1})=\sigma(\mathcal{P}_{1}), so, because \sigma is an orientation, A_{1}<A_{2} implies A_{2}=\sigma(\mathcal{P}_{2})=[\sigma,A](\mathcal{P}_{2}).
Thus, we have proved that [\sigma,A] defines an orientation. Conversely, if \sigma(\mathcal{P}) is not minimal in \sigma(\mathfrak{P}), then there exists some \mathcal{Q}\in\mathfrak{P}\backslash\{\mathcal{P}\} satisfying \sigma(\mathcal{Q})<\sigma(\mathcal{P}). Noticing that [\sigma,A](\mathcal{Q})=\sigma(\mathcal{Q})<\sigma(\mathcal{P}) but [\sigma,A](\mathcal{P})=A\neq\sigma(\mathcal{P}), we deduce that [\sigma,A] does not define an orientation. ∎
As an immediate consequence, we deduce the following statement.
Corollary 2.48.
Let \sigma be an orientation and D\neq\sigma(\mathcal{P}) a sector delimited by some partition \mathcal{P}. If [\sigma,D] defines an orientation, then so does [\sigma,D^{\prime}] for every D^{\prime}\in\mathcal{P}.
The first step in understanding the geometry of C(X,<,\mathfrak{P}) is to understand its geodesics. This is the purpose of Lemma 2.50 below, but before we need a technical preliminary result.
Lemma 2.49.
Let \sigma be an orientation and D_{1},\ldots,D_{n} a collection of sectors delimited by \mathcal{P}_{1},\ldots,\mathcal{P}_{n} respectively, such that [\sigma,D_{1},\ldots,D_{k}] defines an orientation for every 1\leq k\leq n. If \mathcal{P}_{n} is transverse to \mathcal{P}_{1},\ldots,\mathcal{P}_{n1}, then [\sigma,D_{1},\ldots,D_{n}]=[\sigma,D_{n},D_{1},\ldots,D_{n1}] and [\sigma,D_{n},D_{1},\ldots,D_{k}] defines an orientation for every 1\leq k\leq n1.
Proof.
Notice that, for every 1\leq k\leq n1 and \mathcal{P}\in\mathfrak{P}, we have
[\sigma,D_{n},D_{1},\ldots,D_{k}](\mathcal{P})=\left\{\begin{array}[]{cl}[% \sigma,D_{1},\ldots,D_{k}](\mathcal{P})&\text{if}\ \mathcal{P}\neq\mathcal{P}_% {n}\\ D_{n}&\text{if}\ \mathcal{P}=\mathcal{P}_{n}\end{array}\right.=[\sigma,D_{1},% \ldots,D_{k},D_{n}](\mathcal{P}), 
hence [\sigma,D_{n},D_{1},\ldots,D_{k}]=[\sigma,D_{1},\ldots,D_{k},D_{n}]. In particular, if we set \mu_{k}=[\sigma,D_{1},\ldots,D_{k}], it is sufficient to prove that [\mu_{k},D_{n}] defines an orientation to deduce that [\sigma,D_{n},D_{1},\ldots,D_{k}] defines an orientation as well, since \mu_{k} is an orientation by assumption and [\mu_{k},D_{n}]=[\sigma,D_{n},D_{1},\ldots,D_{k}] by our previous observation. Precisely, we want to prove that, if A_{1}\in\mathcal{Q}_{1} and A_{2}\in\mathcal{Q}_{2} satisfy A_{1}<A_{2} and A_{1}=[\mu_{k},D_{n}](\mathcal{Q}_{1}), then A_{2}=[\mu_{k},D_{n}](\mathcal{Q}_{2}). Notice that our assumption implies that \mathcal{Q}_{1} and \mathcal{Q}_{2} are nested. Two cases may happen.
Suppose that \mathcal{Q}_{1}\neq\mathcal{P}_{n}. As a consequence, A_{1}=[\mu_{k},D_{n}](\mathcal{Q}_{1})=\mu_{k}(\mathcal{Q}_{1}), so that A_{1}<A_{2} implies A_{2}=[\sigma,D_{1},\ldots,D_{k}](\mathcal{Q}_{2}). If \mathcal{Q}_{2}\neq\mathcal{P}_{n}, then A_{2}=[\sigma,D_{1},\ldots,D_{k},D_{n}](\mathcal{Q}_{2}), and we are done. So suppose that \mathcal{Q}_{2}=\mathcal{P}_{n}. In particular, since \mathcal{Q}_{2}=\mathcal{P}_{n} is not transverse to \mathcal{Q}_{1}, we deduce that \mathcal{Q}_{1}\neq\mathcal{P}_{i} for every 1\leq i\leq n, hence A_{1}=[\sigma,D_{1},\ldots,D_{n}](\mathcal{Q}_{1}). Since [\sigma,D_{1},\ldots,D_{n}] is an orientation, A_{1}<A_{2} implies that A_{2}=[\sigma,D_{1},\ldots,D_{n}](\mathcal{Q}_{2})=D_{n}=[\sigma,D_{1},\ldots,% D_{k},D_{n}](\mathcal{Q}_{2}).
Suppose that \mathcal{Q}_{1}=\mathcal{P}_{n}. Then A_{1}=[\sigma,D_{1},\ldots,D_{k},D_{n}](\mathcal{Q}_{1})=D_{n}. Noticing that \mathcal{Q}_{1}=\mathcal{P}_{n} is not transverse to \mathcal{Q}_{2}, necessarily \mathcal{Q}_{2}\neq\mathcal{P}_{i} for every 1\leq i\leq n. On the other hand, [\sigma,D_{1},\ldots,D_{n}](\mathcal{Q}_{1})=D_{n}=A_{1}<A_{2} implies A_{2}=[\sigma,D_{1},\ldots,D_{n}](\mathcal{Q}_{2}). Therefore, A_{2}=[\sigma,D_{1},\ldots,D_{n}](\mathcal{Q}_{2})=\sigma(\mathcal{Q}_{2})=[% \sigma,D_{1},\ldots,D_{k},D_{n}](\mathcal{Q}_{2}). This concludes the proof. ∎
Lemma 2.50.
Let \sigma be an orientation and D_{1},\ldots,D_{n} a collection of sectors, delimited by the walls \mathcal{P}_{1},\ldots,\mathcal{P}_{n} respectively, such that
\sigma,\ [\sigma,D_{1}],\ [\sigma,D_{1},D_{2}],\ldots,[\sigma,D_{1},\ldots,D_{% n}] 
defines a path in C(X,\mathfrak{P}). This path is a geodesic if and only if, for every i\neq j, \mathcal{P}_{i}\neq\mathcal{P}_{j} and D_{i}\neq\sigma(\mathcal{P}_{i}).
Proof.
Suppose that there exist some 1\leq i<k\leq n such that \mathcal{P}_{i}=\mathcal{P}_{k}. We choose i and k such that \mathcal{P}_{r}\neq\mathcal{P}_{s} for every 1\leq r<s\leq k1.
Suppose by contradiction that there exists some i<j<k such that \mathcal{P}_{j} and \mathcal{P}_{i} are nested; let A be the sector of \mathcal{P}_{i}=\mathcal{P}_{k} which contains \mathcal{P}_{j}, and B the sector of \mathcal{P}_{j} which contains \mathcal{P}_{i}=\mathcal{P}_{k}. If D_{i}=D_{k}, then the fact that \mathcal{P}_{r}\neq\mathcal{P}_{s} for every 1\leq r<s\leq k1 implies that [\sigma,D_{1},\ldots,D_{k}](\mathcal{P})=[\sigma,D_{1},\ldots,D_{k1}](% \mathcal{P}) for every wall \mathcal{P}\in\mathfrak{P}, hence [\sigma,D_{1},\ldots,D_{k}]=[\sigma,D_{1},\ldots,D_{k1}]. As a consequence, we can shorten our path as
\sigma,\ [\sigma,D_{1}],\ldots,[\sigma,D_{1},\ldots,D_{k1}],\ [\sigma,D_{1},% \ldots,D_{k1},D_{k+1}],\ldots,[\sigma,D_{1},\ldots,D_{k1},D_{k+1},\ldots,D_{% n}]. 
Similarly, if D_{j}=\sigma(\mathcal{P}_{j}), then [\sigma,D_{1},\ldots,D_{j}]=[\sigma,D_{1},\ldots,D_{j1}] so that we are able to shorten out path. The same argument holds if D_{i}=\sigma(\mathcal{P}_{i}). From now on, we suppose that D_{i}\neq D_{k}, D_{i}\neq\sigma(\mathcal{P}_{i}) and D_{j}\neq\sigma(\mathcal{P}_{j}). Now, we distinguish two cases.
Suppose that D_{i}\neq A. Then D_{i}<B with D_{i}=[\sigma,D_{1},\ldots,D_{i}](\mathcal{P}_{i}), hence B=[\sigma,D_{1},\ldots,D_{i}](\mathcal{P}_{j})=\sigma(\mathcal{P}_{j}). On the other hand, D_{j}\neq\sigma(\mathcal{P}_{j})=B implies D_{j}<A, so that we deduce from D_{j}=[\sigma,D_{1},\ldots,D_{j}](\mathcal{P}_{j}) that A=[\sigma,D_{1},\ldots,D_{j}](\mathcal{P}_{i})=D_{i}, a contradiction.
Suppose that D_{i}=A. In particular, D_{k}\neq A so D_{k}<B. On the other hand, we know that D_{k}=[\sigma,D_{1},\ldots,D_{k}](\mathcal{P}_{k}) hence B=[\sigma,D_{1},\ldots,D_{k}](\mathcal{P}_{j})=D_{j}. But \sigma(\mathcal{P}_{j})\neq D_{j}=B implies \sigma(\mathcal{P}_{j})<A, so that we deduce that D_{i}=A=\sigma(\mathcal{P}_{i}), a contradiction.
Thus, we have proved that \mathcal{P}_{k} is transverse to \mathcal{P}_{j} for every i<j<k. As a consequence of Lemma 2.49, we deduce that
[\sigma,D_{1},\ldots,D_{k}]=[\sigma,D_{1},\ldots,D_{i},D_{k},D_{i+1},\ldots,D_% {k1}]=[\sigma,D_{1},\ldots,D_{i1},D_{k},D_{i+1},\ldots,D_{k1}] 
and that [\sigma,D_{1},\ldots,D_{i1},D_{k},D_{i+1},\ldots,D_{i+s}] defines an orientation for every 0\leq s\leq ki. Thus, it is possible to shorten our path by replacing the subsegment
\sigma,\ [\sigma,D_{1}],\ [\sigma,D_{1},D_{2}],\ldots,[\sigma,D_{1},\ldots,D_{% k}] 
with
\sigma,[\sigma,D_{1}],\ldots,[\sigma,D_{1},\ldots,D_{i1}],[\sigma,D_{1},% \ldots,D_{i1},D_{k},],[\sigma,D_{1},\ldots,D_{i1},D_{k},D_{i+1}],\ldots, 
[\sigma,D_{1},\ldots,D_{i1},D_{k},D_{i+1},\ldots,D_{k1}] 
Thus, we have prove that, if there exist 1\leq i,j\leq n such that \mathcal{P}_{i}=\mathcal{P}_{j}, then our path is not a geodesic. Now, suppose that \mathcal{P}_{i}\neq\mathcal{P}_{j} for every i\neq j but \sigma(\mathcal{P}_{i})=D_{i} for some 1\leq i\leq n. Then
[\sigma,D_{1},\ldots,D_{i1}](\mathcal{P}_{i})=\sigma(\mathcal{P}_{i})=D_{i}=[% \sigma,D_{1},\ldots,D_{i}](\mathcal{P}_{i}), 
hence [\sigma,D_{1},\ldots,D_{i1}]=[\sigma,D_{1},\ldots,D_{i}]. Therefore, we can shorten our path as
\sigma,\ [\sigma,D_{1}],\ldots,[\sigma,D_{1},\ldots,D_{i1}],\ [\sigma,D_{1},% \ldots,D_{i1},D_{i+1}],\ldots,[\sigma,D_{1},\ldots,D_{i1},D_{i+1},\ldots,D_{% n}]. 
In particular, our path was not a geodesic.
Conversely, it is clear that the distance between two orientations is at least equal to the number of walls on which they differ. On the other hand, if we suppose that, for every i\neq j, \mathcal{P}_{i}\neq\mathcal{P}_{j} and \mathcal{P}_{i}\neq\sigma(\mathcal{P}_{i}), then \sigma and [\sigma,D_{1},\ldots,D_{n}] differ precisely on \mathcal{P}_{1},\ldots,\mathcal{P}_{n}, so that our path must be a geodesic. ∎
Notice that we have proved, in the last paragraph of the previous proof, that:
Corollary 2.51.
The distance between two vertices of C(X,\mathfrak{P}) is equal to the number of walls on which they differ.
The following lemma will be fundamental in the proof of Proposition 2.46.
Lemma 2.52.
Let \sigma\in C(X,\mathfrak{P}) be an orientation and D_{1},\ldots,D_{n} some sectors such that
\sigma,\ [\sigma,D_{1}],\ [\sigma,D_{1},D_{2}],\ldots,[\sigma,D_{1},\ldots,D_{% n}] 
defines a geodesic in C(X,\mathfrak{P}). Fix some 1\leq i\leq n. If there is no 1\leq j\leq n different from i such that D_{j}<D_{i}, then there exists a permutation \varphi of \{1,\ldots,n\} such that
\sigma,\ [\sigma,D_{\varphi(1)}],\ [\sigma,D_{\varphi(1)},D_{\varphi(2)}],% \ldots,[\sigma,D_{\varphi(1)},\ldots,D_{\varphi(n)}] 
defines a geodesic as well, with \varphi(n)=i.
Proof.
Let \mathcal{P}_{1},\ldots,\mathcal{P}_{n} denote the walls delimiting the sectors D_{1},\ldots,D_{n} respectively. We argue by induction over n. If n=1, there is nothing to prove. From now on, suppose that n\geq 2.
First, we claim that there exists some k\neq i such that \sigma(\mathcal{P}_{k}) is minimal in \sigma(\mathfrak{P}). Suppose by contradiction that this is not the case. Because we already know that \sigma(\mathcal{P}_{1}) is minimal in \sigma(\mathfrak{P}), since [\sigma,D_{1}] defines an orientation, necessarily i=1. Then, notice that, for every wall \mathcal{Q}\in\mathfrak{P} and any 1\leq j\leq n, \sigma(\mathcal{Q})<\sigma(\mathcal{P}_{j}) implies \mathcal{Q}\in\{\mathcal{P}_{1},\ldots,\mathcal{P}_{n}\}. Indeed, if \mathcal{Q}\notin\{\mathcal{P}_{1},\ldots,\mathcal{P}_{n}\}, we deduce from [\sigma,D_{1},\ldots,D_{n}](\mathcal{Q})=\sigma(\mathcal{Q})<\sigma(\mathcal{P% }_{j}) that \sigma(\mathcal{P}_{j})=[\sigma,D_{1},\ldots,D_{n}](\mathcal{P}_{j})=D_{j}, which is impossible according to Lemma 2.50. As a consequence, since for every 2\leq j\leq n the sector \sigma(\mathcal{P}_{j}) is not minimal in \sigma(\mathfrak{P}), we have \sigma(\mathcal{P}_{1})<\sigma(\mathcal{P}_{j}).
In particular, \mathcal{P}_{1} and \mathcal{P}_{j} must be nested; let A_{j} denote the sector delimited by \mathcal{P}_{1} which contains \mathcal{P}_{j}, and notice that \sigma(\mathcal{P}_{j}) is the sector delimited by \mathcal{P}_{j} which contains \mathcal{P}_{1}. If A_{j}=D_{1}, then, because we know that \sigma(\mathcal{P}_{j})\neq D_{j} thanks to Lemma 2.50, we deduce that D_{j}<A_{j}=D_{1}, which is impossible. Otherwise, if A_{j}\neq D_{1}, we have D_{1}<\sigma(\mathcal{P}_{j}), so that [\sigma,D_{1},\ldots,D_{n}](\mathcal{P}_{1})=D_{1}<\sigma(\mathcal{P}_{j}) implies that \sigma(\mathcal{P}_{j})=[\sigma,D_{1},\ldots,D_{n}](\mathcal{P}_{j})=D_{j}, which is impossible according to Lemma 2.50.
Thus, we have proved that there exists some k\neq i such that \sigma(\mathcal{P}_{k}) is minimal in \sigma(\mathfrak{P}). As a consequence, \sigma^{\prime}=[\sigma,D_{k}] is an orientation. To conclude the proof by applying our induction hypothesis, it is sufficient to show that the path
\sigma^{\prime},[\sigma^{\prime},D_{1}],\ldots,[\sigma^{\prime},D_{1},\ldots,D% _{k1}],[\sigma^{\prime},D_{1},\ldots,D_{k1},D_{k+1}],\ldots,[\sigma^{\prime}% ,D_{1},\ldots,D_{k1},D_{k+1},\ldots,D_{n}] 
defines a geodesic in C(X,<,\mathfrak{P}). We first need to verify that it defines a path in C(X,<,\mathfrak{P}). Notice that, for every 0\leq r\leq nk and every \mathcal{P}\in\mathfrak{P},
\begin{array}[]{lcl}[\sigma^{\prime},D_{1},\ldots,D_{k1},D_{k+1},\ldots,D_{k+% r}](\mathcal{P})&=&\left\{\begin{array}[]{cl}\sigma^{\prime}(\mathcal{P})=% \sigma(\mathcal{P})&\text{if}\ \mathcal{P}\notin\{\mathcal{P}_{1},\ldots,% \mathcal{P}_{k+r}\}\\ D_{j}&\text{if}\ \mathcal{P}=\mathcal{P}_{j},\ j\in\{1,\ldots,k+r\}\backslash% \{k\}\\ \sigma^{\prime}(\mathcal{P}_{k})=D_{k}&\text{if}\ \mathcal{P}=\mathcal{P}_{k}% \end{array}\right.\\ \\ &=&[\sigma,D_{1},\ldots,D_{k+r}](\mathcal{P}),\end{array} 
so [\sigma^{\prime},D_{1},\ldots,D_{k1},D_{k+1},\ldots,D_{k+r}] is an orientation. Similarly, if i<k, we have [\sigma^{\prime},D_{1},\ldots,D_{i}]=[\sigma,D_{1},\ldots,D_{i},D_{k}]. Because we know that [\sigma,D_{1},\ldots,D_{i}] is an orientation, it is sufficient to show that [\sigma,D_{1},\ldots,D_{i}](\mathcal{P}_{k})=\sigma(\mathcal{P}_{k}) is minimal in [\sigma,D_{1},\ldots,D_{i}](\mathfrak{P}) in order to deduce that [\sigma^{\prime},D_{1},\ldots,D_{i}] defines an orientation. So let \mathcal{Q}\in\mathfrak{P} be a wall satisfying [\sigma,D_{1},\ldots,D_{i}](\mathcal{Q})<\sigma(\mathcal{P}_{k}), and suppose by contradiction that \mathcal{Q}\neq\mathcal{P}_{k}. Because \sigma(\mathcal{P}_{k}) is minimal in \sigma(\mathfrak{P}), necessarily \mathcal{Q}\in\{\mathcal{P}_{1},\ldots,\mathcal{P}_{i}\}, so that D_{j}<\sigma(\mathcal{P}_{k}) for some 1\leq j\leq i. Now, we deduce from [\sigma,D_{1},\ldots,D_{k}](\mathcal{P}_{j})=D_{j}<\sigma(\mathcal{P}_{k}) that \sigma(\mathcal{P}_{k})=[\sigma,D_{1},\ldots,D_{k}](\mathcal{P}_{k})=D_{k}, contradicting Lemma 2.50.
Finally, we can apply Lemma 2.50 to conclude that our path is a geodesic. Indeed, we know that \mathcal{P}_{i}\neq\mathcal{P}_{j} for every i\neq j, and, for every i\in\{1,\ldots,n\}\backslash\{k\}, we have \sigma^{\prime}(\mathcal{P}_{i})=\sigma(\mathcal{P}_{i})\neq D_{i}. This concludes the proof. ∎
It is clear from the definition of C(X,<,\mathcal{P}) that any of its edges is naturally labelled by a wall of \mathfrak{P}. Our last step before proving Proposition 2.46 is to understand how behave these labels in the triangles and squares of C(X,<,\mathfrak{P}).
Lemma 2.53.
The edges of a triangle in C(X,<,\mathfrak{P}) are labelled by the same wall.
Proof.
Let \alpha,\beta,\gamma\in C(X,<,\mathfrak{P}) be three pairwise adjacent vertices. Let \mathcal{A,B,C} denote the walls labelling the edges (\beta,\gamma),(\alpha,\gamma),(\alpha,\beta) respectively. If \mathcal{A}\neq\mathcal{B}, then \alpha and \beta differ on two walls, but this is impossible according to Corollary 2.51 since d(\alpha,\beta)=1. Therefore, \mathcal{A}=\mathcal{B}. Similarly, we show that \mathcal{B}=\mathcal{C}, concluding the proof. ∎
Lemma 2.54.
Two opposite edges of some square in C(X,<,\mathfrak{P}) are labelled by the same wall of \mathfrak{P}. Moreover, the two walls labelling the edges of some induced square are transverse.
Proof.
Let (\sigma,\mu,\nu,\xi) be a square in C(X,<,\mathfrak{P}). If \mu and \xi, or \sigma and \nu, are adjacent, then we deduce from Lemma 2.53 that all the walls labelling the edges of our square are identical. From now on, we will suppose that our square is induced. As a consequence, \sigma,\mu,\nu and \sigma,\xi,\nu define two geodesics between \sigma and \nu. According to Lemma 2.50, there exist A_{1}\in\mathcal{P}_{1}, A_{2}\in\mathcal{P}_{2}, B_{1}\in\mathcal{Q}_{1} and B_{2}\in\mathcal{Q}_{2} such that \mu=[\sigma,B_{1}], \nu=[\sigma,B_{1},B_{2}], \xi=[\sigma,A_{1}] and \nu=[\sigma,A_{1},A_{2}], with \mathcal{P}_{1}\neq\mathcal{P}_{2}, \mathcal{Q}_{1}\neq\mathcal{Q}_{2}, A_{1}\neq\sigma(\mathcal{P}_{1}), A_{2}\neq\sigma(\mathcal{P}_{2}), B_{1}\neq\sigma(\mathcal{Q}_{1}) and B_{2}\neq\sigma(\mathcal{Q}_{2}). In particular, we deduce that \sigma and \nu=[\sigma,A_{1},A_{2}]=[\sigma,B_{1},B_{2}] differ on the walls \{\mathcal{P}_{1},\mathcal{P}_{2},\mathcal{Q}_{1},\mathcal{Q}_{2}\}. On the other hand, we know that they differ on only two walls according to Corollary 2.51 since d(\sigma,\nu)=2, so we deduce that \{\mathcal{P}_{1},\mathcal{P}_{2}\}=\{\mathcal{Q}_{1},\mathcal{Q}_{2}\}. Next, we know similarly that \mu and \xi must differ on exactly two walls, so we deduce from
[\sigma,A_{1}](\mathcal{P})=\left\{\begin{array}[]{cl}\sigma(\mathcal{P})&% \text{if}\ \mathcal{P}\neq\mathcal{P}_{1}\\ A_{1}&\text{if}\ \mathcal{P}=\mathcal{P}_{1}\end{array}\right.\ \text{and}\ [% \sigma,B_{1}](\mathcal{P})=\left\{\begin{array}[]{cl}\sigma(\mathcal{P})&\text% {if}\ \mathcal{P}\neq\mathcal{Q}_{1}\\ B_{1}&\text{if}\ \mathcal{P}=\mathcal{Q}_{1}\end{array}\right. 
that \mathcal{P}_{1} and \mathcal{Q}_{1} must be different. Therefore, \mathcal{Q}_{1}=\mathcal{P}_{2} and \mathcal{Q}_{2}=\mathcal{P}_{1}. As a consequence, notice that
B_{1}=[\sigma,B_{1},B_{2}](\mathcal{P}_{2})=\nu(\mathcal{P}_{2})=[\sigma,A_{1}% ,A_{2}](\mathcal{P}_{2})=A_{2}, 
and similarly B_{2}=A_{1}. Thus, [\sigma,A_{1},A_{2}]=[\sigma,A_{2},A_{1}].
Now, suppose by contradiction that \mathcal{P}_{1} and \mathcal{P}_{2} are nested, and let C_{1} (resp. C_{2}) denote the sector delimited by \mathcal{P}_{1} (resp. \mathcal{P}_{2}) which contains \mathcal{P}_{2} (resp. \mathcal{P}_{1}). Two cases may happen.
Suppose that A_{1}=C_{1} and A_{2}=C_{2}. Then \sigma(\mathcal{P}_{1})\neq A_{1} implies \sigma(\mathcal{P}_{1})<C_{2}=A_{2}, hence \sigma(\mathcal{P}_{2})=A_{2}, a contradiction.
Suppose that either A_{1}\neq C_{1} or A_{2}\neq C_{2}. Because the two possibilities are symmetric, say that A_{2}\neq C_{2}. We deduce from [\sigma,A_{1},A_{2}](\mathcal{P}_{2})=A_{2}<C_{1} that C_{1}=[\sigma,A_{1},A_{2}](\mathcal{P}_{1})=A_{1}. As a consequence, \sigma(\mathcal{P}_{1})\neq A_{1}=C_{1} implies \sigma(\mathcal{P}_{1})<C_{2}, hence C_{2}=\sigma(\mathcal{P}_{2}). But \sigma(\mathcal{P}_{1})<\sigma(\mathcal{P}_{2}) is impossible since \sigma(\mathcal{P}_{2}) must be minimal in \sigma(\mathfrak{P}). ∎
Proof of Proposition 2.46..
First, we want to prove that C(X,<,\mathfrak{P}) satisfies the triangle condition. Let \mu,\nu\in C(X,<,\mathfrak{P}) be two adjacent orientations and \sigma\in C(X,<,\mathfrak{P}) a third orientation satisfying d(\sigma,\mu)=d(\sigma,\nu)=k. Because d(\mu,\nu)=1, the orientations \mu and \nu differ on a single wall \mathcal{P}_{0}, say \nu=[\mu,D] where D is a sector delimited by \mathcal{P}_{0}. We write \mu=[\sigma,D_{1},\ldots,D_{k}], where
\sigma,\ [\sigma,D_{1}],\ [\sigma,D_{1},D_{2}],\ldots,[\sigma,D_{1},\ldots,D_{% k}] 
defines a geodesic between \sigma and \mu. Notice that there exists some 1\leq i\leq k such that the underlying wall of D_{i} is \mathcal{P}_{0}, since otherwise the path
\sigma,\ [\sigma,D_{1}],\ldots,[\sigma,D_{1},\ldots,D_{k}],\ [\sigma,D_{1},% \ldots,D_{k},D] 
would define a geodesic of length k+1 between \sigma and \nu, according to Lemma 2.50. Notice that
\nu=[\mu,D]=[\sigma,D_{1},\ldots,D_{k},D]. 
In particular, because D_{i} and D have the same underlying wall (and \mathcal{P}_{0} is not the underlying wall of any other D_{j}), D_{i}=[\sigma,D_{1},\ldots,D_{k}](\mathcal{P}_{0}) must be minimal in [\sigma,D_{1},\ldots,D_{k}](\mathfrak{P}) according to Lemma 2.47, so that D_{j}\subset D_{i} for no j\neq i. It follows from Lemma 2.52 that there exists a permutation \varphi of \{1,\ldots,k\} satisfying \varphi(k)=i such that
\sigma,\ [\sigma,D_{\varphi(1)}],\ [\sigma,D_{\varphi(1)},D_{\varphi(2)}],% \ldots,[\sigma,D_{\varphi(1)},\ldots,D_{\varphi(k)}] 
defines a geodesic between \sigma and \mu. Let \xi=[\sigma,D_{\varphi(1)},\ldots,D_{\varphi(k1)}]. Then d(\sigma,\xi)=d(\sigma,\mu)1=k1, and [\xi,D_{i}]=\mu, and [\xi,D]=\nu. Therefore, \xi is the orientation we are looking for.
Now, we want to prove that C(X,<,\mathfrak{P}) satisfies the quadrangle condition. Let \alpha,\beta\in C(X,<,\mathfrak{P}) be two orientations both adjacent to a third one \gamma\in C(X,<,\mathfrak{P}), and let \sigma\in C(X,<,\mathfrak{P}) be a last orientation satisfying d(\sigma,\alpha)=d(\sigma,\beta)=k, d(\sigma,\gamma)=k+1. Because we already know that the triangle condition holds, we will suppose that \alpha and \beta are not adjacent, ie., d(\alpha,\beta)=2. Fix some geodesic
\sigma,\ [\sigma,D_{1}],\ [\sigma,D_{1},D_{2}],\ldots,[\sigma,D_{1},\ldots,D_{% k}]=\alpha 
between \sigma and \alpha. Let S_{1},S_{2} be some sectors such that \gamma=[\alpha,S_{1}] and \beta=[\gamma,S_{2}], and let \mathcal{P}_{1} and \mathcal{P}_{2} denote the walls delimiting S_{1} and S_{2} respectively. Because d(\sigma,\gamma)=k+1, the concatenation of a geodesic between \sigma and \alpha with the edge between \alpha and \gamma must be a geodesic, so we deduce from Lemma 2.50 that, for every 1\leq i\leq k, the underlying wall of D_{i} is different from \mathcal{P}_{1}. Now, concatenating this geodesic with the edge (\gamma,\beta) produces a path which cannot be a geodesic, so that Lemma 2.50 implies that \mathcal{P}_{2} is the underlying wall of S_{1} or of some D_{i}. Notice that, because \alpha and \beta are not adjacent, \mathcal{P}_{1}\neq\mathcal{P}_{2}, so there exists some 1\leq j\leq k such that the underlying wall of D_{j} is \mathcal{P}_{2}. Since we know that
\beta=[\gamma,S_{2}]=[\alpha,S_{1},S_{2}]=[\sigma,D_{1},\ldots,D_{k},S_{1},S_{% 2}], 
and that \beta and \gamma are two orientations, necessarily D_{j}=[\sigma,D_{1},\ldots,D_{k},S_{1}](\mathcal{P}_{2}) is minimal in [\sigma,D_{1},\ldots,D_{k},S_{1}](\mathfrak{P}). As a consequence, if D_{j} is not minimal in [\sigma,D_{1},\ldots,D_{k}](\mathfrak{P}), necessarily
\sigma(\mathcal{P}_{1})=[\sigma,D_{1},\ldots,D_{k}](\mathcal{P}_{1})<D_{j}, 
hence D_{j}=\sigma(\mathcal{Q}_{j}), where \mathcal{Q}_{j} denotes the underlying wall of D_{j}, which contradicts Lemma 2.50. Therefore, D_{j} is minimal in [\sigma,D_{1},\ldots,D_{k}](\mathfrak{P}), so that D_{j}>D_{i} does not hold for any 1\leq i\leq k. According to Lemma 2.52, there exists a permutation \varphi of \{1,\ldots,k\} satisfying \varphi(k)=j such that
\sigma,\ [\sigma,D_{\varphi(1)}],\ [\sigma,D_{\varphi(1)},D_{\varphi(2)}],% \ldots,[\sigma,D_{\varphi(1)},\ldots,D_{\varphi(k)}] 
defines a geodesic between \sigma and \beta. Set \xi=[\sigma,D_{\varphi(1)},\ldots,D_{\varphi(k1)}]. Notice that d(\sigma,\xi)=k1 and that \xi is adjacent to \alpha since [\xi,D_{j}]=\alpha. To conclude that \xi is the orientation we are looking for, it is sufficient to show that \xi and \beta are adjacent.
We know that d(\sigma,\gamma)=k+1, ie., \sigma and \gamma differ on k+1 walls, so, because \beta=[\gamma,S_{2}], \sigma(\mathcal{P}_{2})\neq S_{2} would imply that \sigma and \beta differ on k+1 walls, contradicting the fact that d(\sigma,\beta)=k. Therefore, \sigma(\mathcal{P}_{2})=S_{2}. As a consequence, [\xi,S_{1}] is equal to \beta=[\alpha,S_{1},S_{2}]=[\xi,D_{j},S_{1},S_{2}]. Indeed, these two orientations may only differ on \mathcal{P}_{1} and \mathcal{P}_{2}, but
[\xi,S_{1}](\mathcal{P}_{1})=S_{1}=[\xi,D_{j},S_{1},S_{2}](\mathcal{P}_{1}), 
and
[\xi,S_{1}](\mathcal{P}_{2})=\xi(\mathcal{P}_{2})=\sigma(\mathcal{P}_{2})=S_{2% }=[\xi,D_{j},S_{1},S_{2}](\mathcal{P}_{2}). 
This concludes the proof of the quadrangle condition.
Finally, we need to verify that C(X,<,\mathfrak{P}) does not contain induced subgraphs isomorphic to K_{4}^{} or K_{2,3}. If Y\subset X is a subgraph isomorphic to K_{4}^{}, it follows from Lemma 2.53 that all its edges are labelled by the same wall, which implies that all its vertices must be pairwise adjacent; in particular, Y is not an induced subgraph. Next, if Y is a subgraph isomorphic to K_{2,3}, we deduce from Lemma 2.54 that there exist two nonadjacent vertices of Y which share a common neighbor along two edges labelled by the same wall; necessarily, these two vertices have to be adjacent in X, so that Y is not an induced subgraph. This concludes the proof that any connected component of the quasicubulation C(X,<,\mathfrak{P}) is a quasimedian graph. ∎
It is worth noticing that C(X,<,\mathfrak{P}) is in general not connected. Even worse, it may happen that no natural choice of a connected component is possible. The example to keep in mind is the following. Let S be an infinite set. We consider the partition \mathfrak{P} of the power set 2^{S} containing the walls
\{\{A\subset S\mid x\in A\},\ \{A\subset S\mid x\notin A\}\} 
for every x\in S. The quasicubulation C(2^{S},\subset,\mathfrak{P}) is naturally isometric to the graph whose vertices are the sequences which belong to \{0,1\}^{S} and whose edges link two sequences which differ on a single coordinate, and its connected components are all isomorphic. Nevertheless, we show below that some specific connected components, if they exist, are more strongly related to the initial popset than others.
Definition 2.55.
Let (X,<,\mathfrak{P}) be a popset. An orientation \sigma is wellfounded if, for every wall \mathcal{P}\in\mathfrak{P}, the set \{\mathcal{Q}\in\mathfrak{P}\mid\sigma(\mathcal{Q})<\sigma(\mathcal{P})\} is finite.
Unfortunately, a popset does not always admit a wellfounded orientation; see [Rol98, Example 9.7]. Nevertheless, following respectively [Rol98, Theorem 9.6] and [Rol98, Proposition 9.4], it can be proved that a wellfounded orientation exists whenever our popset (X,<,\mathfrak{P}) is countable and discrete (ie., for every A,B\in X, the set \{C\mid A<C<B\} is finite) or whenever it is discrete and \omegadimensional (ie., every wall \mathcal{P}\in\mathfrak{P} is finite and there does not exist an infinite collection of pairwise transverse walls). Moreover, in the latter case, the wellfounded orientations correspond precisely to the orientations satisfying the descending chain condition, as introduced in [Sag14] for pocsets.
Theorem 2.56.
Let (X,<,\mathfrak{P}) be a popset admitting a wellfounded orientation, and let Y denote a connected component of the quasicubulation C(X,<,\mathfrak{P}) which contains such an orientation. Then Y is a quasimedian graph, and there is a natural bijection between the walls of \mathfrak{P} and the hyperplanes of Y which respects transversality and tangency.
The following definition gives a precise meaning of tangent walls, extending the notion of tangent hyperplanes in CAT(0) cube complexes.
Definition 2.57.
Let (X,<,\mathfrak{P}) be a popset. Two walls \mathcal{P}_{1},\mathcal{P}_{2}\in\mathfrak{P} are tangent if they are nested and if there do not exist A_{1}\in\mathcal{P}_{1}, A_{2}\in\mathcal{P}_{2}, \mathcal{P}\in\mathfrak{P} and A\in\mathcal{P} such that A_{1}<A<A_{2} or A_{2}<A<A_{1}.
Proof of Theorem 2.56..
We already know that Y is a quasimedian graph thanks to Proposition 2.46.
Notice that Lemma 2.53 and Lemma 2.54 imply that the edges of a given hyperplane of C(X,<,\mathfrak{P}) are labelled by the same wall of \mathfrak{P}, so that the hyperplanes of C(X,<,\mathfrak{P}) are naturally labelled by the walls of \mathfrak{P}.
We claim that two distinct hyperplanes of C(X,<,\mathfrak{P}) are labelled by distinct walls of \mathfrak{P}. More precisely, we will prove that two edges labelled by the same wall of \mathfrak{P} are dual to the same hyperplane.
Let e,f^{\prime} be two edges such that e=(\mu,[\mu,A]) and f^{\prime}=(\nu^{\prime},[\nu^{\prime},B]) for some A,B\in\mathcal{P} and \mathcal{P}\in\mathfrak{P}. Notice that, setting \nu=[\nu^{\prime},\mu(\mathcal{P})], the edge f=(\nu,[\nu,A]) is dual to the same hyperplane as f^{\prime}, since the edges
f^{\prime}=(\nu^{\prime},[\nu^{\prime},B]),\ (\nu^{\prime},[\nu^{\prime},\mu(% \mathcal{P})]),\ ([\nu^{\prime},\mu(\mathcal{P})],[\nu^{\prime},\mu(\mathcal{P% }),A])=f 
successively belong to the same triangle. Thus, it is sufficient to prove that e and f are dual to the same hyperplane. Let
\mu,\ [\mu,D_{1}],\ [\mu,D_{1},D_{2}],\ldots,[\mu,D_{1},\ldots,D_{n}]=\nu 
be a geodesic between \mu and \nu. For convenience, let \mathcal{P}_{i} denote the underlying wall of D_{i} for 1\leq i\leq n. According to Lemma 2.50, \mathcal{P}_{i}\neq\mathcal{P}_{j} for every i\neq j; moreover, because \mu(\mathcal{P})=\nu(\mathcal{P}), necessarily \mathcal{P}\neq\mathcal{P}_{i} for every i.
We argue by induction on n. If n=0, then e=f and the hyperplanes dual to e and f are obviously the same. From now on, suppose that n\geq 1 and set \sigma=[\mu,D_{1},\ldots,D_{n1}]. Let \mathcal{Q}\in\mathfrak{P} satisfy \sigma(\mathcal{Q})<\sigma(\mathcal{P}). If \mathcal{Q}\neq\mathcal{P}_{n}, then
\nu(\mathcal{Q})=[\sigma,D_{n}](\mathcal{Q})=\sigma(\mathcal{Q})<\sigma(% \mathcal{P})=[\sigma,D_{n}](\mathcal{P})=\nu(\mathcal{P}). 
Because we already know that \nu(\mathcal{P}) is minimal in \nu(\mathfrak{P}), we deduce that \mathcal{Q}=\mathcal{P}. On the other hand,
\sigma(\mathcal{P}_{n})=[\mu,D_{1},\ldots,D_{n1}](\mathcal{P}_{n})=\mu(% \mathcal{P}_{n})\nless\mu(\mathcal{P})=[\mu,D_{1},\ldots,D_{n1}](\mathcal{P})% =\sigma(\mathcal{P}) 
since we already know that \mu(\mathcal{P}) is minimal in \mu(\mathfrak{P}). Therefore, \sigma(\mathcal{P}) must be minimal in \sigma(\mathfrak{P}), so that [\sigma,A] defines an orientation. Now, notice that, for every \mathcal{Q}\notin\{\mathcal{P},\mathcal{P}_{n}\}, we have [\sigma,A](\mathcal{Q})=\sigma(\mathcal{Q}) and
[\nu,A](\mathcal{Q})=\nu(\mathcal{Q})=[\sigma,D_{n}](\mathcal{Q})=\sigma(% \mathcal{Q}); 
and [\sigma,A](\mathcal{P})=A=[\nu,A](\mathcal{P}); and finally, using Lemma 2.50,
[\sigma,A](\mathcal{P}_{n})=\mu(\mathcal{P}_{n})\neq D_{n}=\nu(\mathcal{P}_{n}% )=[\nu,A](\mathcal{P}_{n}). 
Therefore, [\sigma,A] and [\nu,A] differ on a single wall of \mathfrak{P}, so that they must be adjacent. Thus, we have proved that the edges (\sigma,[\sigma,A]) and f=(\nu,[\nu,A]) are opposite sides of some square; on the other hand, our induction hypothesis implies that e and (\sigma,[\sigma,A]) are dual to the same hyperplane. A fortiori, e and f, and so e and f^{\prime}, are dual to the same hyperplane, concluding the proof of our claim.
We claim that any wall \mathcal{P}\in\mathfrak{P} labels some edge of Y. Let \sigma denote a wellfounded orientation which belongs to Y. Because \sigma is wellfounded, the set
I=\{\mathcal{Q}\mid\sigma(\mathcal{Q})<\sigma(\mathcal{P})\}\backslash\{% \mathcal{P}\} 
is finite, say I=\{\mathcal{Q}_{1},\ldots,\mathcal{Q}_{n}\}. Notice that, for every 1\leq i\leq n, \sigma(\mathcal{Q}_{i})<\sigma(\mathcal{P}) implies that \mathcal{Q}_{i} and \mathcal{P} are nested; let A_{i} denote the sector delimited by \mathcal{Q}_{i} which contains \mathcal{P}. Notice that, for every 1\leq i\leq n, \sigma(\mathcal{P}) is the sector delimited by \mathcal{P} which contains \mathcal{Q}_{i}.
Setting \mu=[\sigma,A_{1},\ldots,A_{n}], we want to prove that \mu is an orientation. So let B_{1}\in\mathcal{R}_{1} and B_{2}\in\mathcal{R}_{2} be two sectors respectively delimited by two walls \mathcal{R}_{1},\mathcal{R}_{2}, satisfying B_{1}<B_{2} and B_{1}=\mu(\mathcal{R}_{1}). Our goal is to show that B_{2}=\mu(\mathcal{R}_{2}). We distinguish two cases.
Suppose that \mathcal{R}_{1}\notin\{\mathcal{Q}_{1},\ldots,\mathcal{Q}_{n}\}. Then we deduce from \sigma(\mathcal{R}_{1})=\mu(\mathcal{R}_{1})=B_{1}<B_{2} that B_{2}=\sigma(\mathcal{R}_{2}). If \mathcal{R}_{2}\notin\{\mathcal{Q}_{1},\ldots,\mathcal{Q}_{n}\}, we conclude that B_{2}=\sigma(\mathcal{R}_{2})=\mu(\mathcal{R}_{2}). Otherwise, say \mathcal{R}_{2}=\mathcal{Q}_{i}, we deduce from \sigma(\mathcal{R}_{1})=B_{1}<B_{2}=\sigma(\mathcal{R}_{2})=\sigma(\mathcal{Q}% _{i})<\sigma(\mathcal{P}), hence \mathcal{R}_{1}\in\{\mathcal{Q}_{1},\ldots,\mathcal{Q}_{n}\}, a contradiction.
Suppose that \mathcal{R}_{1}=\mathcal{Q}_{i} for some 1\leq i\leq n. Notice that A_{i}=\mu(\mathcal{Q}_{i})=\mu(\mathcal{R}_{1})=B_{1}<B_{2}. Therefore, for every D\in\mathcal{P}\backslash\{\sigma(\mathcal{P})\}, we have D<A_{i}<B_{2}, so that B_{2} is the sector of \mathcal{R}_{2} which contains \mathcal{P}. If \mathcal{R}_{2}=\mathcal{Q}_{j} for some 1\leq j\leq n, this means that B_{2}=A_{j}, hence B_{2}=A_{j}=\mu(\mathcal{Q}_{j})=\mu(\mathcal{R}_{2}). Now, suppose that \mathcal{R}_{2}\notin\{\mathcal{Q}_{1},\ldots,\mathcal{Q}_{n}\}, ie., \sigma(\mathcal{R}_{2})\nless\sigma(\mathcal{P}). Therefore, \sigma(\mathcal{R}_{2}) must be the sector of \mathcal{R}_{2} which contains \mathcal{P}, hence B_{2}=\sigma(\mathcal{R}_{2})=\mu(\mathcal{R}_{2}).
Thus, we have proved that \mu is an orientation. Now, we notice that \mu(\mathcal{P})=\sigma(\mathcal{P}) is minimal in \mu(\mathfrak{P}). Indeed, if \mathcal{Q}\in\mathfrak{P} is a wall satisfying \mu(\mathcal{Q})<\sigma(\mathcal{P}), then either \mathcal{Q}\notin\{\mathcal{Q}_{1},\ldots,\mathcal{Q}_{n}\}, so that \sigma(\mathcal{Q})=\mu(\mathcal{Q})<\sigma(\mathcal{P}) which implies \mathcal{Q}=\mathcal{P}; or \mathcal{Q}=\mathcal{Q}_{i} for some 1\leq i\leq n, so that A_{i}=\mu(\mathcal{Q}_{i})=\mu(\mathcal{Q})<\sigma(\mathcal{P}), which is impossible since A_{i} and \sigma(\mathcal{P}) are not <comparable according to Lemma 2.42. Therefore, if we fix some D\in\mathcal{P}\backslash\{\sigma(\mathcal{P})\}, then the two orientations \mu and [\mu,D] define two vertices of Y linked by an edge, which is clearly labelled by \mathcal{P}. This concludes the proof of our claim.
We claim that two hyperplanes J_{1},J_{2} of C(X,<,\mathfrak{P}) are transverse if and only if the walls \mathcal{P}_{1},\mathcal{P}_{2} which label them are transverse as well.
Suppose that J_{1} and J_{2} are transverse. Then there exists some square whose dual hyperplanes are J_{1} and J_{2}. We deduce from Lemma 2.54 that \mathcal{P}_{1} and \mathcal{P}_{2} must be transverse. Conversely, suppose that \mathcal{P}_{1} and \mathcal{P}_{2} are transverse. Let \sigma be a wellfounded orientation which belongs to Y. Using exactly the same argument as above, we show that the set
I=\{\mathcal{Q}\mid\sigma(\mathcal{Q})<\sigma(\mathcal{P}_{1})\}\backslash\{% \mathcal{P}_{1}\} 
is finite, say I=\{\mathcal{A}_{1},\ldots,\mathcal{A}_{n}\}, that \mu=[\sigma,A_{1},\ldots,A_{n}] is an orientation if A_{i} denotes the sector of \mathcal{A}_{i} containing \mathcal{P}_{1} for every 1\leq i\leq n, and finally that \sigma(\mathcal{P}_{1}) is minimal in \mu(\mathfrak{P}). Applying this argument once again, we show that the set
J=\{\mathcal{Q}\mid\mu(\mathcal{Q})<\mu(\mathcal{P}_{2})\}\backslash\{\mathcal% {P}_{2}\} 
is finite, say J=\{\mathcal{B}_{1},\ldots,\mathcal{B}_{m}\}, that \nu=[\mu,B_{1},\ldots,B_{m}] is an orientation if B_{i} denotes the sector of \mathcal{B}_{i} containing \mathcal{P}_{2} for every 1\leq i\leq m, and finally that \mu(\mathcal{P}_{2}) is minimal in \nu(\mathfrak{P}). Notice that, because \mathcal{P}_{1} and \mathcal{P}_{2} are transverse, they do not belong to I\cup J. As a consequence, \nu(\mathcal{P}_{2})=\mu(\mathcal{P}_{2}) is minimal in \nu(\mathfrak{P}). Now, we want to prove that \nu(\mathcal{P}_{1})=\sigma(\mathcal{P}_{1}) is minimal in \nu(\mathfrak{P}).
First, we notice that I\cap J=\emptyset. Indeed, suppose by contradiction that there exists a wall \mathcal{Q}\in\mathfrak{P}\backslash\{\mathcal{P}_{1},\mathcal{P}_{2}\} satisfying \sigma(\mathcal{Q})<\sigma(\mathcal{P}_{1}) and \mu(\mathcal{Q})<\mu(\mathcal{P}_{2}). Let D\in\mathcal{P}_{1}\backslash\{\sigma(\mathcal{P}_{1})\}. Because \mu(\mathcal{Q}) is the sector delimited by \mathcal{Q} which contains \mathcal{P}_{1}, we deduce that D<\mu(\mathcal{Q})<\mu(\mathcal{P}_{2}), so that \mathcal{P}_{1} and \mathcal{P}_{2} must be nested, a contradiction.
Let \mathcal{Q}\in\mathfrak{P} be a wall satisfying \nu(\mathcal{Q})<\sigma(\mathcal{P}_{1}). Our goal is to prove that \mathcal{Q}=\mathcal{P}_{1}. If \mathcal{Q}\notin J, then we deduce from \mu(\mathcal{Q})=\nu(\mathcal{Q})<\sigma(\mathcal{P}_{1}) that \mathcal{Q}=\mathcal{P}_{1} because we already know that \sigma(\mathcal{P}_{1}) is minimal in \mu(\mathfrak{P}). Now, suppose that Q\in J. Let D\in\mathcal{P}_{1}\backslash\{\nu(\mathcal{P}_{1})\}. From \nu(\mathcal{Q})<\sigma(\mathcal{P}_{1}), we deduce that \nu(\mathcal{P}_{1})=\sigma(\mathcal{P}_{1}) is the sector delimited by \mathcal{P}_{1} which contains \mathcal{Q}, so that D must be included into the sector delimited by \mathcal{Q} which contains \mathcal{P}_{1}. On the other hand, we observed that I\cap J=\emptyset, so that \mathcal{Q}\notin I, hence \mu(\mathcal{Q})=\sigma(\mathcal{Q})\nless\sigma(\mathcal{P}_{1})=\nu(\mathcal% {P}_{1}). A fortiori, \mu(\mathcal{Q}) is the sector delimited by \mathcal{Q} which contains \mathcal{P}_{1}, hence D<\mu(\mathcal{Q}). But \mathcal{Q}\in J implies \mu(\mathcal{Q})<\mu(\mathcal{P}_{2}), so that D<\mu(\mathcal{Q})<\mu(\mathcal{P}_{2}). We conclude that \mathcal{P}_{1} and \mathcal{P}_{2} must be nested, a contradiction. Thus, we have proved that \sigma(\mathcal{P}_{1}) is minimal in \nu(\mathfrak{P}).
We conclude that, if we fix two sectors D_{1}\in\mathcal{P}_{1}\backslash\{\nu(\mathcal{P}_{1})\} and D_{2}\in\mathcal{P}_{2}\backslash\{\nu(\mathcal{P}_{2})\}, then [\nu,D_{1}] and [\nu,D_{2}] are two orientations. Moreover, since \mathcal{P}_{1} and \mathcal{P}_{2} are transverse, we deduce that \nu(\mathcal{P}_{1}) is minimal in [\nu,D_{2}](\mathcal{P}), so that [\nu,D_{1},D_{2}] is an orientation as well. Finally, the four orientations \nu, [\nu,D_{1}], [\nu,D_{2}] and [\nu,D_{1},D_{2}]=[\nu,D_{2},D_{1}] define a square in Y whose dual hyperplanes are J_{1} and J_{2}. A fortiori, J_{1} and J_{2} must be transverse.
Finally, we want to prove that two hyperplanes of C(X,<,\mathfrak{P}) are tangent if and only if the walls labelling them are tangent as well.
First of all, let us notice that, if J is a hyperplane of C(X,<,\mathfrak{P}) labelled by some wall \mathcal{P}\in\mathfrak{P}, then the sectors delimited by J are precisely the \{\sigma\mid\sigma(\mathcal{P})=A\}, where A\in\mathcal{P}. Indeed, fix two orientations \mu,\nu and some geodesic
\mu,\ [\mu,A_{1}],\ [\mu,A_{1},A_{2}],\ldots,[\mu,A_{1},\ldots,A_{n}]=\nu 
between them. If \mathcal{P}_{1},\ldots,\mathcal{P}_{n} are the walls underlying A_{1},\ldots,A_{n} respectively, we deduce from Lemma 2.50 that the walls on which \mu and \nu differ are precisely \mathcal{P}_{1},\ldots,\mathcal{P}_{n}. On the other hand, \mu and \nu belong to the same sector delimited by J if and only if J intersects this geodesic, which is equivalent to \mathcal{P}=\mathcal{P}_{i} for some 1\leq i\leq n. Therefore, \mu and \nu belong to the same sector delimited by J if and only if \mu(\mathcal{P})=\nu(\mathcal{P}). This proves our claim.
As a consequence, we are able to prove that, if \mathcal{P}_{1} and \mathcal{P}_{2} are two walls which are not tangent, then the associated hyperplanes J_{1} and J_{2} respectively are not tangent as well. Indeed, if \mathcal{P}_{1} and \mathcal{P}_{2} are not tangent, either \mathcal{P}_{1} and \mathcal{P}_{2} are transverse or there exist A_{1}\in\mathcal{P}_{1}, A_{2}\in\mathcal{P}_{2}, \mathcal{P}\in\mathfrak{P} and A\in\mathcal{P} such that A_{1}<A<A_{2} (up to switching \mathcal{P}_{1} and \mathcal{P}_{2}). In the former case, we know that J_{1} and J_{2} must be transverse, so that they cannot be tangent. In the latter case, we deduce that S_{1}\subset S\subset S_{2}, where S_{1}=\{\sigma\mid\sigma(\mathcal{P}_{1})=A_{1}\} is a sector delimited by J_{1}, S_{2}=\{\sigma\mid\sigma(\mathcal{P}_{2})=A_{2}\} a sector delimited by J_{2} and S=\{\sigma\mid\sigma(\mathcal{P})=A\} a sector delimited by J, which proves that J_{1} and J_{2} are not tangent. Indeed, if \sigma\in S_{1}, then \sigma(\mathcal{P}_{1})=A_{1}<A implies that \sigma(\mathcal{P})=A, ie., \sigma\in S; and similarly, if \sigma\in S, then \sigma(\mathcal{P})=A<A_{2} implies that \sigma(\mathcal{P}_{2})=A_{2}, ie., \sigma\in S_{2}.
Conversely, we claim that, if S_{1}\subsetneq S_{2} are two sectors of C(X,<,\mathfrak{P}) such that S_{1}=\{\sigma\mid\sigma(\mathcal{P}_{1})=A_{1}\} and S_{2}=\{\sigma\mid\sigma(\mathcal{P}_{2})=A_{2}\} for some \mathcal{P}_{1},\mathcal{P}_{2}\in\mathfrak{P}, A_{1}\in\mathcal{P}_{1} and A_{2}\in\mathcal{P}_{2}, then A_{1}<A_{2}. For this purpose, notice that \mathcal{P}_{1} and \mathcal{P}_{2} must be nested and let B_{1} (resp. B_{2}) denote the sector of \mathcal{P}_{1} (resp. \mathcal{P}_{2}) containing \mathcal{P}_{2} (resp. \mathcal{P}_{1}). Fix three orientations \sigma\in S_{1}, \mu\in S_{2}\backslash S_{1} and \nu\notin S_{2}. We distinguish three cases.

•
Suppose that A_{1}\neq B_{1}. Then \sigma(\mathcal{P}_{1})=A_{1}<B_{2}, which implies that B_{2}=\sigma(\mathcal{P}_{2})=A_{2}. Therefore, A_{1}<B_{2}=A_{2}.

•
Suppose that A_{1}=B_{1} and A_{2}=B_{2}. Then \nu(\mathcal{P}_{2})\neq A_{2}=B_{2} implies that \nu(\mathcal{P}_{2})<B_{1}=A_{1}, hence A_{1}=\nu(\mathcal{P}_{1}), which is impossible since \nu\notin S_{1}.

•
Suppose that A_{1}=B_{1} but A_{2}\neq B_{2}. Then \mu(\mathcal{P}_{2})=A_{2}\neq B_{2} implies that \mu(\mathcal{P}_{2})<B_{1}=A_{1}, hence A_{1}=\mu(\mathcal{P}_{1}), which is impossible since \mu\notin S_{1}.
This concludes the proof of our claim.
As a consequence, if \mathcal{P}_{1} and \mathcal{P}_{2} are two walls labelling two hyperplanes J_{1} and J_{2} which are not tangent, then \mathcal{P}_{1} and \mathcal{P}_{2} cannot be tangent. Indeed, if J_{1} and J_{2} are not tangent, either they are transverse or there exists a third hyperplane J delimiting some sector S satisfying S_{1}\subset S\subset S_{2} for some sectors S_{1},S_{2} delimited by J_{1},J_{2} respectively. In the former case, we know that \mathcal{P}_{1} and \mathcal{P}_{2} must be transverse as well, so that they cannot be transverse. In the latter case, if \mathcal{P} denote the wall labelling J, we deduce from our previous claim that there exist A_{1}\in\mathcal{P}_{1}, A_{2}\in\mathcal{P}_{2} and A\in\mathcal{P} satisfying A_{1}<A<A_{2}. A fortiori, \mathcal{P}_{1} and \mathcal{P}_{2} are not tangent. ∎
Remark 2.58.
Suppose that a group G acts on X by preserving the order < and the partition \mathfrak{P}. Then G naturally acts on the quasicubulation C(X,<,\mathfrak{P}) by isometries. Unfortunately, if Y denotes a connected component of C(X,<,\mathfrak{P}) which contains a wellfounded orientation, then Y is not necessarily Ginvariant. Nevertheless, if this is the case, then the bijection between the walls of \mathfrak{P} and the hyperplanes of Y produced by the previous theorem is Gequivariant.
Besides cubulating pocsets, a related method to construct CAT(0) cube complexes is cubulating spaces with walls; see [Nic04, CN05b]. We also generalize this construction. Before defining what we call a space of partitions, we need to introduce some terminology. Fix some set X and some collection of partitions \mathfrak{P}. Elements of \mathfrak{P} will be referred to as partitions and elements of a partition will be referred to as sectors. Two elements \mathcal{P}_{1},\mathcal{P}_{2} are

•
indistinguishable if they represent the same partition of X, otherwise they are distinguishable;

•
nested if there exist two sectors A_{1}\in\mathcal{P}_{1} and A_{2}\in\mathcal{P}_{2} such that D\subset A_{1} for every D\in\mathcal{P}_{2}\backslash\{A_{2}\} and D\subset A_{2} for every D\in\mathcal{P}_{1}\backslash\{A_{1}\}.
Finally, we say that a partition \mathcal{P}\in\mathfrak{P} separates two points of X if they belong to different sectors delimited by \mathcal{P}.
Definition 2.59.
A space with partitions (X,\mathfrak{P}) is the data of a set X and a collection of partitions \mathfrak{P} satisfying the following conditions:

•
every \mathcal{P}\in\mathfrak{P} satisfies \#\mathcal{P}\geq 2 and \emptyset\notin\mathcal{P};

•
for every distinguishable partitions \mathcal{P}_{1},\mathcal{P}_{2}\in\mathfrak{P}, if there exist two sectors A_{1}\in\mathcal{P}_{1}, A_{2}\in\mathcal{P}_{2} such that A_{1}\subset A_{2}, then \mathcal{P}_{1} and \mathcal{P}_{2} are nested.

•
two points of X are separated by finitely many partitions of \mathfrak{P}.
The third assumption allows us to define the pseudodistance d_{\mathfrak{P}} on X, counting the number of partitions separating two points of X. A quasimedian graph X is naturally a space with partitions, without indistinguishable partitions: for the set of partitions \mathfrak{P}, consider the sectordecompositions induced by the hyperplanes of X. In this case, the associated pseudodistance coincides with the initial distance defined on X.
We associate to any space with partitions (X,\mathfrak{P}) a popset in the following way. Let \mathfrak{D} denote the set of all the sectors delimited by the partitions of \mathfrak{P}; by convention, a sector delimited by two distinct partitions is counted twice. In particular, \mathfrak{P} defines a partition of \mathfrak{D}. The order < we define on \mathfrak{D} is: for every sectors D_{1},D_{2}\in\mathfrak{D}, we set D_{1}<D_{2} if D_{1}\subset D_{2} and if the underlying partitions of D_{1} and D_{2} are distinguishable. Then it essentially follows from the definition of spaces with partitions that (\mathfrak{D},<,\mathfrak{P}) is a popset. An orientation of (\mathfrak{D},<,\mathfrak{P}) can be thought of as a map which chooses, for each partition of \mathfrak{P}, a sector it delimits in such a way that, whenever \mathcal{P}_{1} and \mathcal{P}_{2} are two distinguishable partitions delimiting respectively two sectors A_{1},A_{2} satisfying A_{1}<A_{2}, then A_{1}=\sigma(\mathcal{P}_{1}) implies A_{2}=\sigma(\mathcal{P}_{2}). A simple example of a space with partitions and the quasicubulation of its associated popset is given in Figure 4.
If x\in X, we can define the orientation \sigma_{x} associating to any partition of \mathfrak{P} the sector which contains x. Such an orientation is called a principal orientation. These orientations will allow us to choose a canonical connected component of our quasicubulation.
Lemma 2.60.
Any principal orientation defines a wellfounded orientation.
Proof.
Let x\in X be a point and let \sigma_{x} denote the associated principal orientation. Fixing some partition \mathcal{P}\in\mathfrak{P}, we want to prove that the set \{\mathcal{Q}\in\mathfrak{P}\mid\sigma_{x}(\mathcal{Q})<\sigma_{x}(\mathcal{P})\} is finite. Because \#\mathcal{P}\geq 2 and \emptyset\notin\mathcal{P}, there exists a point y\in X which does not belong to \sigma_{x}(\mathcal{P}). Notice that, if \mathcal{Q} belongs to our set, then x\in\sigma_{x}(\mathcal{Q}) but y\notin\sigma_{x}(\mathcal{Q}). Therefore, \mathcal{Q} separates x and y. Since there exist only finitely many partitions of \mathfrak{P} separating two given points of X, this concludes the proof. ∎
Lemma 2.61.
Two principal orientations differ only on finitely many partitions.
Proof.
Let x,y\in X be two points and let \sigma_{x},\sigma_{y} denote the associated principal orientations. If \mathcal{P}\in\mathfrak{P} is a partition satisfying \sigma_{x}(\mathcal{P})\neq\sigma_{y}(\mathcal{P}), then x and y must belong to different sectors delimited by \mathcal{P}, ie., \mathcal{P} separates x and y. Because there exist only finitely many partitions separating two given points of X, we conclude that \sigma_{x} and \sigma_{y} differ on only finitely many partitions. ∎
Thus, there is a canonical choice of a connected component of the quasicubulation of the popset associated to a space with partitions (X,\mathfrak{P}). Namely, this is the connected component which contains all the principal orientations, which we denote by C(X,\mathfrak{P}). It is worth noticing that, if a group G acts on X leaving \mathfrak{P} invariant, then G naturally acts on C(X,\mathfrak{P}) by isometries, since the set of principal orientations is Ginvariant (indeed, if x\in X and g\in G, then g\cdot\sigma_{x}=\sigma_{g\cdot x}). Moreover, according to the following lemma, the canonical map (X,d_{\mathfrak{P}})\hookrightarrow C(X,\mathfrak{P}) defined by x\mapsto\sigma_{x} is a (pseudo)isometric embedding.
Lemma 2.62.
For every x,y\in X, we have d_{C(X,\mathfrak{P})}(\sigma_{x},\sigma_{y})=d_{\mathfrak{P}}(x,y).
Proof.
The partitions on which \sigma_{x} and \sigma_{y} differ are precisely the partitions separating x and y. Therefore, our equality follows from Corollary 2.51. ∎
Important examples of spaces with partitions, as already mentionned, are quasimedian graphs themselves. Our next result characterize their quasicubulations.
Proposition 2.63.
Let X be a quasimedian graph. The quasicubulation of X, viewed as a space with partitions, is isometric to X.
Our proposition will be essentially a consequence of the following lemma.
Lemma 2.64.
Let (X,\mathfrak{P}) be the space with partitions canonically associated to a quasimedian graph X. Any orientation \sigma\in C(X,\mathfrak{P}) is principal.
Proof.
For convenience, we will identify \mathfrak{P} with the set of the hyperplanes of X. It is worth noticing that two hyperplanes are transverse as elements of \mathfrak{P} if and only if they are transverse as hyperplanes of X. Fix an arbitrary vertex x\in X, and let \sigma_{x} denote its associated principal orientation. Because \sigma\in C(X,\mathfrak{P}), there exist finitely many hyperplanes J_{1},\ldots,J_{n} of X on which \sigma and \sigma_{x} differ. Let y denote the projection of x onto C=\bigcap\limits_{i=1}^{n}\sigma(J_{i}) (which is non empty because \sigma is an orientation, thanks to Helly’s property 2.8). Notice that J_{1},\ldots,J_{n} separate x and y. We claim that \sigma=\sigma_{y}.
Let J be a hyperplane of X. If J does not separate x and y, then \sigma(J)=\sigma_{x}(J)=\sigma_{y}(J). Otherwise, J separates x from C according to Lemma 2.34. On the other hand, because \sigma is an orientation, \sigma(J)\cap\sigma(J_{i})\neq\emptyset for every 1\leq i\leq n, so that Helly’s property 2.8 implies that \sigma(J)\cap C\neq\emptyset. As a consequence, \sigma(J) must be the sector delimited by J which contains C; hence \sigma(J)=\sigma_{y}(J).
Thus, \sigma=\sigma_{y}. We have proved that \sigma is a principal orientation. ∎
Proof of Proposition 2.63..
Suppose that X is a quasimedian graph and that \mathfrak{P} is the collection of the sectordecompositions its the hyperplanes. Consider the map
\Sigma:\left\{\begin{array}[]{ccc}X&\longrightarrow&C(X,\mathfrak{P})\\ x&\longmapsto&\sigma_{x}\end{array}\right. 
which sends each vertex of X to the associated principal orientation. Because, for every vertices x,y\in X, the distance between \sigma_{x} and \sigma_{y} in C(X,\mathfrak{P}) is equal to the number of hyperplanes separating x and y in X, which is equal to the distance between x and y in X, we know that \Sigma defines an isometric embedding X\hookrightarrow C(X,\mathfrak{P}). Moreover, the surjectivity of \Sigma follows from Lemma 2.64, so X and C(X,\mathfrak{P}) are isometric. ∎
Proposition 2.63 will be useful to prove some results on quasimedian graphs. Let us mention a first application.
Corollary 2.65.
Let X be a quasimedian graph and Y\subset X a subgraph. Then Y is finite if and only if the set of the sectors separating two vertices of Y is finite.
Proof.
The implication is clear, so suppose the set of the sectors separating two vertices of Y is finite. Fix some vertex x\in Y. For every y\in Y, let D(y) denote the set of the sectors containing y but not x. Set \mathcal{D}=\bigcup\limits_{y\in Y}D(y). Notice that any sector of \mathcal{D} separates two vertices of Y, hence \#\mathcal{D}<+\infty because Y is finite and that there exist only finitely many sectors separating two given vertices of Y. On the other hand, the map
\left\{\begin{array}[]{ccc}Y&\longrightarrow&2^{\mathcal{D}}\\ y&\longmapsto&D(y)\end{array}\right. 
is injective, since \sigma_{y}=[\sigma_{x},D(y)] for every y\in Y. Therefore, the finiteness of \mathcal{D} implies the finiteness of Y. ∎
Remark 2.66.
Pocsets are popsets where all the walls have cardinality two, and spaces with walls are spaces with partitions whose partitions all have cardinality two. In these cases, our quasicubulation coincides with the usual cubulations, as introduced respectively in [Sag14] and [Nic04, CN05b], and so produces a median graph. Alternatively, we can notice that the quasimedian graph produced by quasicubulating a pocset is trianglefree. Indeed, as a consequence of Lemma 2.53, the edges of a triangle are labelled by a common wall, so that this wall necessarily delimits at least three sectors. Therefore, it follows from Corollary 2.92 that the quasimedian graph we obtain turns out to be a median graph.
2.5 Gated hulls
In this section, we notice that gated subgraphs of quasimedian graphs behave like combinatorially convex subcomplexes of CAT(0) cube complexes; see for instance [Hag08, Paragraph 2.3].
Definition 2.67.
Let X be a graph and S\subset X a set of vertices. The gated hull of S is the smallest gated subgraph of X containing S; alternatively, this is the intersection of all the gated subgraphs of X which contain S.
The main result of this section is the following proposition.
Proposition 2.68.
Let X be a quasimedian graph and S\subset X a subset. The hyperplanes of the gated hull Y of S are precisely the restrictions to Y of the hyperplanes of X separating two vertices of S. Moreover, two hyperplanes of Y are transverse if and only if their extensions are transverse in X.
We begin by proving the following lemma, which is wellknown for CAT(0) cube complexes.
Lemma 2.69.
The gated hull of a subset S\subset X is the intersection of all the sectors containing S.
Proof.
Let Y denote the intersection of all the sectors containing S. Because the intersection of gated subgraphs is gated, it follows from Corollary 2.22 that Y is gated. Now, we want to prove that, for any gated subgraph Z containing S, necessarily Y\subset Z. Suppose by contradiction that there exists a vertex x\in Y\backslash Z. In particular, there exists a hyperplane J separating x from its projection onto Z, and we deduce from Lemma 2.34 that J separates x from Z. A fortiori, since S\subset Z, J separates x and S, producing a sector containing S but not x\in Y. This contradicts the definition of Y. ∎
Proof of Proposition 2.68..
It is clear that a hyperplane of Y extends to a hyperplane of X. However, we need to verify that two distinct hyperplanes of Y extends to two distinct hyperplanes of X. More precisely, we want to prove that, if e,f are two edges of Y which are dual to the same hyperplane J in X, then they are dual to the same hyperplane in Y. In fact, using the isomorphism given by Lemma 2.28, it is clear that the gated hull of e\cup f in N(J), which must be included into Y, contains a sequence of edges between e and f such that two consecutive edges either are two opposite sides of the same square or belong to the same triangle. A fortiori, e and f are dual to the same hyperplane in Y.
Now, in order to conclude the proof of the first assertion in our proposition, it is sufficient to show that a hyperplane of X separates two vertices of Y if and only if it separates two vertices of S. If J is a hyperplane which does not separate two vertices of S, then S must be included into a sector delimited by J, and it follows from Lemma 2.69 that J is disjoint from Y. Conversely, since S\subset Y, it is obvious that any hyperplane separating two vertices of S separates two vertices of Y.
Let J_{1},J_{2} be two hyperplanes of Y, which we identify with their extensions to X for convenience. Clearly, if they are transverse in Y then they are transverse in X. Conversely, we suppose that J_{1},J_{2} are transverse in X and we want to prove that they are transverse in Y as well.
Fix two edges (a,b),(c,d)\subset Y dual to J_{1},J_{2} respectively. Let a^{\prime},b^{\prime} denote the projections onto N(J_{2}) of a,b respectively. Since (a,b) is not dual to J_{2}, the vertices a,b belong to the same sector delimited by J_{2} according to Lemma 2.20, so that a^{\prime},b^{\prime} belong to the same connected component \partial_{1} of \partial J_{2}. Let \partial_{2} be a connected component of \partial J_{2} which contains either c or d and which is different from \partial_{1}. Let a^{\prime\prime},b^{\prime\prime} denote the projections onto \partial_{2} of a,b respectively. Notice that, according to Corollary 2.40, a^{\prime\prime},b^{\prime\prime} are also the projections onto \partial_{2} of a^{\prime},b^{\prime} respectively, so that a^{\prime} and b^{\prime} must be adjacent to a^{\prime\prime} and b^{\prime\prime} respectively. Then, because J_{1} separates a and b and intersects N(J_{2}), Proposition 2.33 implies that a^{\prime}\neq b^{\prime} and a^{\prime\prime}\neq b^{\prime\prime}; on the other hand, as a consequence of Corollary 2.35, a^{\prime} and b^{\prime}, and a^{\prime\prime} and b^{\prime\prime}, must be adjacent. Therefore, the vertices a^{\prime},a^{\prime\prime},b^{\prime},b^{\prime\prime} define some square Q whose hyperplanes are J_{1} and J_{2}. Now, since a^{\prime} (resp. b^{\prime}) belongs to a geodesic between a and c (resp. between b and c), the convexity of Y implies a^{\prime}\in Y (resp. b^{\prime}\in Y). Similarly, a^{\prime\prime} (resp. b^{\prime\prime}) belongs to a geodesic between a and \{c,d\}\cap\partial_{2} (resp. between b and \{c,d\}\cap\partial_{2}), hence a^{\prime\prime}\in Y (resp. b^{\prime\prime}\in Y). Therefore, Q\subset Y, and we conclude that J_{1} and J_{2} are transverse in Y. ∎
In CAT(0) cube complexes, the convex hull of a finite set is finite. Although we mentionned that gated subgraphs in quasimedian graphs are the analogue of convex subcomplexes in CAT(0) cube complexes, this assertion does not hold in the world of quasimedian graphs, ie., the gated hull of a finite set may be infinite. Indeed, the gated hull of two vertices which belong to an infinite clique will be the whole infinite clique. Nevertheless, some finiteness property can be deduced from the previous proposition.
Definition 2.70.
A quasimedian graph is cubically finite if it contains finitely many hyperplanes.
The terminology is justified by the observation that a CAT(0) cube complex is finite if and only if it contains finitely many hyperplanes. Corollary 2.81 will give an equivalent characterization of cubically finite quasimedian graphs.
Corollary 2.71.
The gated hull of a finite subset is cubically finite.
Proof.
Let S\subset X be a finite subset, and let Y denote its gated hull. We deduce from Proposition 2.68 that there exists a bijection between the hyperplanes of Y and the hyperplanes of X which separate at least two vertices of S. Because S is finite and that only finitely many hyperplanes separate two given vertices, we conclude that Y contains finitely many hyperplanes, ie., it is cubically finite. ∎
2.6 Prisms
Recall that a weak Cartesian product of graphs is a connected component of the Cartesian product of these graphs.
Definition 2.72.
Let X be a graph. An induced subgraph Y\subset X is a prism if it is a weak Cartesian product of cliques of X. In particular, X is a prism if it is isomorphic to a weak Cartesian product of complete graphs.
In this section, we want to show that cubes in CAT(0) cube complexes are replaced with prisms in quasimedian graphs. Our first main result is the following proposition.
Proposition 2.73.
Let X be a quasimedian graph. If J_{1},\ldots,J_{n} is a collection of pairwise transverse hyperplanes, then X contains a prism whose dual hyperplanes are precisely J_{1},\ldots,J_{n}.
We begin by proving the following lemma.
Lemma 2.74.
Let X be a quasimedian graph whose hyperplanes are pairwise transverse. Let \mathfrak{J} denote its collection of hyperplanes, and for every J\in\mathfrak{J}, let K(J) be the complete graph whose vertices are the sectors delimited by J. Then X is isometric to a connected component of the Cartesian product \prod\limits_{J\in\mathfrak{J}}K(J). In particular, X is a prism.
Proof.
According to Proposition 2.63, we can identify X with the quasicubulation C(X,\mathfrak{P}) of the canonical space of partitions (X,\mathfrak{P}) associated to X. In particular, the vertices of \prod\limits_{J\in\mathfrak{J}}K(J) can be naturally thought of maps associating to any hyperplane one of its sectors. If x\in X is a base vertex, we denote by P the connected component of \prod\limits_{J\in\mathfrak{J}}K(J) which contains \sigma_{x}, ie., the connected component containing all the principal orientations, since two principal orientations differ only on finitely many hyperplanes according to Lemma 2.61. Now, we defined
\Psi:\left\{\begin{array}[]{ccc}X&\longrightarrow&P\\ y&\longmapsto&\sigma_{y}\end{array}\right.. 
Notice that \Psi defines an isometric embedding, since, for every vertices y,z\in X,
\begin{array}[]{lcl}d(\Psi(y),\Psi(z))&=&d(\sigma_{y},\sigma_{z})\\ &=&\#\{\text{hyperplanes on which $\sigma_{y}$ and $\sigma_{z}$ differ}\}\\ &=&\#\{\text{hyperplanes separating $y$ and $z$}\}\\ &=&d(y,z)\end{array} 
Now, we want to prove that \Psi is surjective. Let p\in P be a vertex. Because p and \sigma_{x} belong to the same connected componenent of \prod\limits_{J\in\mathfrak{J}}K(J), p differ from \sigma_{x} on finitely many coordinates p_{1},\ldots,p_{k}. Set
\sigma=[\sigma_{x},p_{1},\ldots,p_{k}]. 
We claim that \sigma defines an orientation. Notice that, if \mu is an orientation and D a sector, because the hyperplanes of X are pairwise transverse, no two elements of (\mu(\mathfrak{P}),\subset) are comparable, so that any element turns out to be minimal; It follows from Lemma 2.47 that [\mu,D] always defines an orientation. By applying this observation successively to \sigma_{x}, to [\sigma_{x},p_{1}], to [\sigma_{x},p_{1},p_{2}], and so on, it follows that \sigma defines an orientation.
Thus, we deduce from Lemma 2.64 that \sigma=\sigma_{y} for some vertex y\in X. By construction, we have \Psi(y)=p, which concludes the proof. ∎
Proof of Proposition 2.73..
Because J_{1},\ldots,J_{n} are pairwise transverse, we deduce from Helly’s property 2.8 that \bigcap\limits_{i=1}^{n}N(J_{i})\neq\emptyset. Let x\in\bigcap\limits_{i=1}^{n}N(J_{i}) be a vertex. For every 1\leq i\leq n, fix some sector D_{i} delimited by J_{i} which does not contain x, and let x_{i}\in D_{i} be a vertex adjacent to x (such a vertex exists because x\in N(J_{i})). Finally, let C denote the gated hull of \{x,x_{1},\ldots,x_{n}\}. We claim that C is the prism we are looking for. In fact, we are going to prove the following statement:
Fact 2.75.
If J_{1},\ldots,J_{n} is a collection of pairwise transverse hyperplanes, x\in\bigcap\limits_{i=1}^{n}N(J_{i}) a vertex, and, for every 1\leq i\leq n, x_{i} a vertex adjacent to x separated from it by J_{i}, then the gated hull of \{x,x_{1},\ldots,x_{n}\} is a prism whose dual hyperplanes are J_{1},\ldots,J_{n}.
According to Proposition 2.68, the hyperplanes of C are naturally identified to the hyperplanes of X separating x and x_{i} or x_{i} and x_{j} for some 1\leq i,j\leq n. In particular, J_{1},\ldots,J_{n} are hyperplanes of C. Now, notice that, if for some 1\leq i,j\leq n the vertices x_{i} and x_{j} are adjacent, then J_{i}=J_{j}, which is impossible; so d(x_{i},x_{j})=2. As a consequence, if i\neq j, the concatenation (x_{i},x)\cup(x,x_{j}) is a geodesic, so that any hyperplane separating x_{i} and x_{j} must separate either x and x_{i} or x and x_{j}. We deduce that the hyperplanes of C are precisely J_{1},\ldots,J_{n}.
Moreover, we know from Proposition 2.68 that two hyperplanes are transverse in a gated subgraph if and only if they are transverse in the whole graph. Therefore, the hyperplanes of C are pairwise transverse, and it follows from Lemma 2.74 that C is a prism on its own right. We conclude that C is a prism of X because, as C is a gated subgraph, it has to be a induced and a clique in C must be a clique in X. ∎
Remark 2.76.
Proposition 2.73 does not hold for infinite collections of pairwise transverse hyperplanes. For instance, there exist CAT(0) cube complexes containing infinite collections of pairwise transverse hyperplanes but containing no infinite cubes. See [HW12, Figure6] for a simple explicit example, or [Gen16c, Section A.1] for examples admitting interesting group actions.
It is worth noticing that, during the proof of Lemma 2.74, the hypothesis on the hyperplanes of X was not used to prove that \Psi is an isometric embedding. Therefore, any quasimedian graph embeds isometrically into a prism. Let us give an alternative proof of this observation, which will be useful later.
Lemma 2.77.
Let X be a quasimedian graph. Let \mathfrak{J} denote the collection of the hyperplanes of X, and, for every J\in\mathfrak{J}, fix a clique C(J) dual to J. The map
x\mapsto\left(\mathrm{proj}_{C(J)}(x)\right)_{J\in\mathfrak{J}} 
defines an isometric embedding X\hookrightarrow\prod\limits_{J\in\mathfrak{J}}C(J).
Proof.
Let x,y\in X be two vertices. For every J\in\mathfrak{J}, we know that \mathrm{proj}_{C(J)}(x)\neq\mathrm{proj}_{C(J)}(y) if and only if J separates x and y. Therefore, the number of coordinates on which \left(\mathrm{proj}_{C(J)}(x)\right)_{J\in\mathfrak{J}} and \left(\mathrm{proj}_{C(J)}(y)\right)_{J\in\mathfrak{J}} differ is equal to the number of hyperplanes separating x and y. This precisely means that our map is an isometric embedding, concluding the proof. ∎
In particular, Proposition 2.73 justifies the following definition.
Definition 2.78.
The cubical dimension of a quasimedian graph X, denoted by \dim_{\square}X, is the maximal number of pairwise intersecting hyperplanes.
We conclude this section by our second and last main result.
Proposition 2.79.
Let X be a quasimedian graph of finite cubical dimension. The map \mathfrak{J}\mapsto\bigcap\limits_{J\in\mathfrak{J}}N(J) defines an \mathrm{Aut}(X)equivariant bijection between the maximal collections of pairwise transverse hyperplanes and the maximal prisms of X.
To prove this proposition, the following lemma will be needed.
Lemma 2.80.
In a quasimedian graph, a prism is gated.
Proof.
Let X be a quasimedian graph. Let \mathcal{S} be a collection of cliques of X and C\subset X a prism which is a weak Cartesian product of the cliques in \mathcal{S}.
First, we claim that any edge e of C belongs to a clique of C which is a clique of X. This is clear if there exists some S\in\mathcal{S} such that e\subset S. Otherwise, C contains a subgraph isomorphic to S\times[0,1], for some S\in\mathcal{S}, such that S\times\{0\}=S and e\subset S\times\{1\}. It follows from Lemma 2.13 that, because S is a clique in X, necessarily S\times\{1\} has to be a clique in X as well. This proves our claim.
Now, we can prove that C contains its triangles. Indeed, if (a,b,c) is a triangle with (a,b)\subset C, then, according to our previous claim, there exists some clique Y of C which is a clique of X and which contains (a,b). Because X does not contain induced subgraphs isomorphic to K_{4}^{}, necessarily c\in Y, hence (a,b,c)\subset Y\subset C.
Finally, we prove that C is locally convex, which is sufficient to conclude that C is gated according to Proposition 2.6. Let (a,b,c,d)\subset X be a square with (a,b)\cup(b,c)\subset C. If a and c are adjacent, we conclude that (a,b,c,d)\subset Y because we already know that C contains its triangles. From now on, we suppose that a and c are not adjacent. Because C is a weak Cartesian product of complete graphs, two adjacent edges generate either a triangle or a square in C, so that there exists a vertex x\in C defining a square (a,b,c,x)\subset C. If d=x, we conclude that (a,b,c,d)\subset C and we are done. Suppose by contradiction that d\neq x. In particular, the vertices a,b,c,d,x define a subgraph isomorphic to K_{2,3}. Because X does not contain induced subgraphs isomorphic to K_{4}^{} and that we supposed that a and c are not adjacent, we deduce that neither b and d, nor b and x, nor d and x, are adjacent, so that we find an induced subgraph in X which is isomorphic to K_{2,3}. ∎
Proof of Proposition 2.79..
Let J_{1},\ldots,J_{n} a maximal collection of pairwise transverse hyperplanes, and set C=\bigcap\limits_{i=1}^{n}N(J_{i}). Notice that, if J is a hyperplane intersecting C, then, for every 1\leq i\leq n, either J=J_{i} or J is transverse to J_{i} since J intersects N(J_{i}). By maximality of our collection, we deduce that no hyperplane different from J_{1},\ldots,J_{n} intersects C; in particular, we deduce from Lemma 2.74 that C is a prism. Conversely, it follows from Fact 2.75 that J_{1},\ldots,J_{n} intersect C, so that the hyperplanes dual to C are precisely J_{1},\ldots,J_{n}. Finally, the maximality of our collection J_{1},\ldots,J_{n} implies that C cannot be strictly contained in a larger prism, so that C is indeed a maximal prism of X.
The injectivity of our application is clear, because the hyperplanes dual to the final maximal prism are precisely the initial collection of hyperplanes.
Now, if C is a maximal prism of X, and if J_{1},\ldots,J_{n} are the hyperplanes dual to C, we claim that C=\bigcap\limits_{i=1}^{n}N(J_{i}), proving that our application is surjective. The inclusion C\subset\bigcap\limits_{i=1}^{n}N(J_{i}) is clear. If we prove that J_{1},\ldots,J_{n} defines a maximal collection of pairwise transverse hyperplanes, then the reverse inclusion will be a consequence of the maximality of C, since we already noticed that \bigcap\limits_{i=1}^{n}N(J_{i}) defines a prism in this case. More precisely, we will prove that, if C is a prism of X and J_{1},\ldots,J_{n} its dual hyperplanes, and if there exists a hyperplane J transverse to J_{1},\ldots,J_{n}, then C is included into a larger prism, and in particular it is not maximal.
Let x\in C and y\in N(J) be two vertices minimizing the distance between C and N(J). Because C is gated according to Lemma 2.80, we apply Lemma 2.36 to deduce that the hyperplanes separating x and y are precisely the hyperplanes separating C and N(J). Let J_{0} denote the hyperplane dual to the first edge of some fixed geodesic \gamma from x to y. In particular, J_{0} separates x and y so that it must separate C and N(J). On the other hand, the hyperplanes J_{1},\ldots,J_{n} intersect both C and N(J), so we deduce that J_{0} has to be transverse to J_{1},\ldots,J_{n}. For every 1\leq i\leq n, let x_{i}\in C be a vertex adjacent to x and separated from it by J_{i}; and let x_{0} denote the vertex of \gamma adjacent to x, so that J_{0} is the hyperplane separating x and x_{0}. If C^{\prime} denote the gated hull of \{x,x_{1},\ldots,x_{n}\} and C^{\prime\prime} the gated hull of \{x,x_{0},x_{1},\ldots,x_{n}\}, we have C\subseteq C^{\prime}\subsetneq C^{\prime\prime}. On the other hand, because J_{0},J_{1},\ldots,J_{n} are pairwise transverse, we deduce from Fact 2.75 that C^{\prime\prime} is a prism. ∎
Corollary 2.81.
Let X be a quasimedian graph. The following assertions are equivalent:

(i)
X is cubically finite;

(ii)
\dim_{\square}(X)<+\infty and X contains finitely many maximal prisms.

(iii)
every prism of X is contained in a maximal prism and X contains finitely many maximal prisms;
Proof.
Suppose that X is cubically finite. Of course, the cubical dimension of X is necessarily finite (and bounded above by the number of hyperplanes of X). As an immediate consequence of Proposition 2.79, it also follows that X contains finitely many maximal prisms. This proves (i)\Rightarrow(ii). Next, suppose (ii). If X contains a prism which is not contained in a maximal prism, then there must exist an increasing sequence of prisms in X, and looking at the hyperplanes dual to this sequence of prisms, we find an infinite collection of pairwise transverse hyperplanes, contradicting the finiteness of the cubical dimension. Therefore, (ii) implies (iii). Finally, suppose (iii). If X is covered by N maximal prisms and if D is the maximal cubical dimension of these prisms, then necessarily X contains at most N\cdot D hyperplanes. A fortiori, X is cubically finite, hence (iii)\Rightarrow(i). ∎
2.7 Quasimedians
Since quasimedian graphs may contain triangles, the median point of a triple of vertices is generally not welldefined. In this section, we prove nevertheless that any triple admits a quasimedian. In fact, quasimedian graphs were essentially defined in [BMW94] in terms of quasimedians, where various equivalent definitions are given, including the definition we use in this paper.
Definition 2.82.
Let X be a graph and x,y,z\in X three vertices. A median triangle of (x,y,z) is a triple of vertices (x^{\prime},y^{\prime},z^{\prime}) satisfying
\left\{\begin{array}[]{l}d(x,y)=d(x,x^{\prime})+d(x^{\prime},y^{\prime})+d(y^{% \prime},y)\\ d(x,z)=d(x,x^{\prime})+d(x^{\prime},z^{\prime})+d(z^{\prime},z)\\ d(y,z)=d(y,y^{\prime})+d(y^{\prime},z^{\prime})+d(z^{\prime},z)\end{array}% \right.. 
The quantity \min(d(x^{\prime},y^{\prime}),d(y^{\prime},z^{\prime}),d(z^{\prime},x^{\prime})) will be referred to as the size of the median triangle. A quasimedian of (x,y,z) is an equilateral median triangle (x^{\prime},y^{\prime},z^{\prime}) (ie., d(x^{\prime},y^{\prime})=d(x^{\prime},z^{\prime})=d(y^{\prime},z^{\prime})) of minimal size.
It is worth noticing that any triple (x,y,z) admits at least one median triangle, namely (x,y,z) itself. Also, if (x^{\prime},y^{\prime},z^{\prime}) denotes a median triangle of some triple (x,y,z), then, for any choice of geodesics [x,x^{\prime}], [x^{\prime},y^{\prime}], [y^{\prime},y] respectively between x and x^{\prime}, x^{\prime} and y^{\prime}, y^{\prime} and y, the concatenation [x,x^{\prime}]\cup[x^{\prime},y^{\prime}]\cup[y^{\prime},y^{\prime}] turns out to be a geodesic. And a similar statement holds for x,x^{\prime},z^{\prime},z and y,y^{\prime},z^{\prime},z.
Definition 2.83.
A graph is median if any triple of vertices admits a unique quasimedian of size zero, which we refer to as the median vertex.
The main result of this section is the following:
Proposition 2.84.
In a quasimedian graph X, any triple of vertices x,y,z admits a unique quasimedian. Moreover, its size is equal to the number of hyperplanes separating the three vertices x,y,z and its gated hull is a prism of finite cubical dimension.
We begin by proving the following technical lemma.
Lemma 2.85.
Let X be a quasimedian graph, \mathcal{S} a finite collection of pairwise intersecting sectors, and x\in X a vertex. Any hyperplane separating x from its projection onto \bigcap\mathcal{S} delimits a sector which contains an element of \mathcal{S}.
Proof.
Let us define inductively a finite sequence of vertices (x_{i}) as follows. Set x_{0}=x. Suppose now that x_{i} is defined for some i\geq 0. If x_{i}\in\bigcap\mathcal{S}, then x_{i} is the last term of our sequence. Otherwise, there exists some S_{i}\in\mathcal{S} such that x_{i}\notin S_{i}, and we define x_{i+1} as the projection of x_{i} onto S_{i}. Let x_{0},\ldots,x_{n}\in X be the sequence defined in this way (and S_{0},\ldots,S_{n1}\in\mathcal{S} the corresponding sequence of sectors). Notice that x_{n}\in\bigcap\mathcal{S}
First of all, let us notice that
Fact 2.86.
For every 0\leq i\leq n1 and 1\leq k\leq ni, the vertex x_{i+k} belongs to S_{i}.
We fix i and argue by induction on k. Because x_{i+1} is by construction the projection of x_{i} onto S_{i}, of course x_{i+1} belongs to S_{i}. Now, suppose that x_{i+k} belongs to S_{i}. As a consequence of Lemma 2.40, we have
\begin{array}[]{lcl}x_{i+k+1}&=&\mathrm{proj}_{S_{i+k}}(x_{i+k})=\mathrm{proj}% _{S_{i+k}}\circ\mathrm{proj}_{S_{i}}(x_{i+k})\\ \\ &=&\mathrm{proj}_{S_{i+k}\cap S_{i}}(x_{i+k})\in S_{i}\end{array} 
which concludes the proof of our fact.
For every 0\leq i\leq n1, fix a geodesic [x_{i},x_{i+1}]. We claim that the concatenation [x_{0},x_{1}]\cup\cdots\cup[x_{n1},x_{n}] is a geodesic. Indeed, if J is a hyperplane intersecting [x_{i},x_{i+1}] for some 0\leq i\leq n1, then, since x_{i+1} is the projection of x_{i} onto S_{i}, it follows from Lemma 2.34 that J separates x_{i} from S_{i}. On the other hand, we know from our previous fact that x_{j} belongs to S_{i} for every j\geq i+1, so that J cannot intersect [x_{i+1},x_{i+2}]\cup\cdots\cup[x_{n1},x_{n}]. This proves that our concatenation turns out to be a geodesic. As a consequence, any hyperplane separating x_{0} and x_{n} must separate x_{i} and x_{i+1} for some 0\leq i\leq n1.
Now, we claim that x_{n} is the projection of x=x_{0} onto \bigcap\mathcal{S}. Let p denote this projection. Because x_{n}\in\bigcap\mathcal{S}, there exists a geodesic between x and x_{n} passing through p, so that any hyperplane separating x_{n} and p must separate x_{n} and x, and finally x_{i} and x_{i+1} for some 0\leq i\leq n1. Because x_{i+1} is the projection of x_{i} onto S_{i}, it follows from Lemma 2.34 that J separates x_{i} from S_{i}. On the other hand, \bigcap\mathcal{S} is included into S_{i}, so it follows that J must be disjoint from \bigcap\mathcal{S}, which is impossible since it separates two of its vertices, namely x_{n} and p. As a consequence, there do not exist hyperplanes separating x_{n} and p, ie., x_{n}=p is the projection of x onto \bigcap\mathcal{S}.
We are finally ready to prove our lemma. So let J be a hyperplane separating x and x_{n}. We know that J must separate x_{i} and x_{i+1} for some 0\leq i\leq n1. And once again because x_{i+1} is the projection of x_{i} onto S_{i}, we deduce from Lemma 2.34 that J separates x_{i} from S_{i}. Therefore, S_{i}\in\mathcal{S} is included into a sector delimited by J, which concludes the proof. ∎
Our previous lemma allows us to prove the following characterisation of median triangles:
Lemma 2.87.
Let X be a quasimedian graph and x,y,z\in X three vertices. Set \mathcal{S} the set of the sectors S satisfying S\cap\{x,y,z\}=2. A triple (x^{\prime},y^{\prime},z^{\prime}) is a median triangle of (x,y,z) if and only if there exists some subcollection \mathcal{S}_{0}\subset\mathcal{S} such that x^{\prime},y^{\prime},z^{\prime} are respectively the projections of x,y,z onto \bigcap\mathcal{S}_{0}.
Proof.
Let \mathcal{S}_{0}\subset\mathcal{S} be a subcollection and let x^{\prime},y^{\prime},z^{\prime} denote respectively the projections of x,y,z onto \bigcap\mathcal{S}_{0}. We want to prove that (x^{\prime},y^{\prime},z^{\prime}) is a median triangle of (x,y,z). By symmetry, it is sufficient to prove that d(x,y)=d(x,x^{\prime})+d(x^{\prime},y^{\prime})+d(y^{\prime},y). Equivalently, fixing some geodesics [x,x^{\prime}], [x^{\prime},y^{\prime}], [y^{\prime},y] respectively between x and x^{\prime}, x^{\prime} and y^{\prime}, y^{\prime} and y, we will show that the concatenation [x,x^{\prime}]\cup[x^{\prime},y^{\prime}]\cup[y^{\prime},y] is a geodesic.
Let J be a hyperplane intersecting [x,x^{\prime}]. It follows from Lemma 2.85 that J delimits a sector S such that S_{0}\subset S for some S_{0}\in\mathcal{S}_{0}. Because S_{0}\in\mathcal{S}_{0}, we know that y\in S_{0}, hence y\in S. Moreover, \bigcap\mathcal{S}_{0}\subset S_{0}\subset S, so x^{\prime},y^{\prime}\in S. Thus, the sector S delimited by J contains x^{\prime},y^{\prime},y, and it follows by convexity of S that [x^{\prime},y^{\prime}]\cup[y^{\prime},y]\subset S. A fortiori, J does not intersect [x^{\prime},y^{\prime}]\cup[y^{\prime},y]. Similarly, one shows that a hyperplane intersecting [y,y^{\prime}] cannot intersect [x,x^{\prime}]\cup[x^{\prime},y^{\prime}]. Therefore, [x,x^{\prime}]\cup[x^{\prime},y^{\prime}]\cup[y^{\prime},y] is a geodesic.
Conversely, suppose that (x^{\prime},y^{\prime},z^{\prime}) is a median triangle of (x,y,z). Let \mathcal{S}_{x} (resp. \mathcal{S}_{y},\mathcal{S}_{z}) denote the collection of the sectors which are delimited by the hyperplanes separating x and x^{\prime} (resp. y and y^{\prime}, z and z^{\prime}) and which contain x^{\prime} (resp. y^{\prime}, z^{\prime}). Finally, set \mathcal{S}_{0}=\mathcal{S}_{x}\cup\mathcal{S}_{y}\cup\mathcal{S}_{z}. We want to show that x^{\prime},y^{\prime},z^{\prime} are respectively the projections of x,y,z onto \bigcap\mathcal{S}_{0}. By symmetry, it is sufficient to prove this statement for x.
First of all, fix some geodesics [x,x^{\prime}],[x^{\prime},y^{\prime}],[x^{\prime},z^{\prime}],[y^{\prime},z^{% \prime}],[y,y^{\prime}],[z,z^{\prime}]. If J is a hyperplane separating x and x^{\prime}, we know that, since a geodesic cannot intersect twice a hyperplane, necessarily J must be disjoint from [x^{\prime},y^{\prime}] (resp. [x^{\prime},z^{\prime}], [y^{\prime},y], [z^{\prime},z]), so that y^{\prime} (resp. z^{\prime}, y, z) must belong to the same sector delimited by J as x^{\prime}. Otherwise saying, we have proved the following statement, which we record for future use:
Fact 2.88.
Given a median triangle (x^{\prime},y^{\prime},z^{\prime}) of a given triple (x,y,z) and a hyperplane J separating x and x^{\prime}, the vertices x^{\prime},y^{\prime},z^{\prime},y,z all belong to a common sector delimited by J.
A similar statement holds for hyperplanes separating y and y^{\prime}, and for those separating z and z^{\prime}, for the same reason. Consequently, the vertices x^{\prime},y^{\prime},z^{\prime} all belong to \bigcap\mathcal{S}_{0}.
Now, let p denote the projection of x onto \bigcap\mathcal{S}_{0}. Since x^{\prime}\in\bigcap\mathcal{S}_{0}, there exists a geodesic between x and x^{\prime} passing through p. If J is a hyperplane separating p and x^{\prime}, then J must separate x and x^{\prime}, so that J delimits a sector S which belongs to \mathcal{S}_{x}\subset\mathcal{S}_{0} (namely, the sector which contains x^{\prime}). But the inclusion \bigcap\mathcal{S}_{0}\subset S implies that J must be disjoint from \bigcap\mathcal{S}_{0}, which is impossible since J separates two vertices of this intersection, namely x^{\prime} and p. Thus, we have proved that no hyperplanes separate x^{\prime} and p, hence x^{\prime}=p. Therefore, x^{\prime} is the projection of x onto \bigcap\mathcal{S}_{0}, concluding the proof. ∎
Proof of Proposition 2.84..
Denote by \mathcal{S} the set of the sectors S satisfying S\cap\{x,y,z\}=2. We claim that, if x^{\prime},y^{\prime},z^{\prime} denote respectively the projections of x,y,z onto \bigcap\mathcal{S}, then (x^{\prime},y^{\prime},z^{\prime}) is a quasimedian of (x,y,z). First of all, we know that this is a median triangle according to Lemma 2.87.
Now, we claim that a hyperplane J separates x^{\prime} and y^{\prime} if and only if it separates x,y,z. Suppose first that J separates x^{\prime} and y^{\prime}. Because (x^{\prime},y^{\prime},z^{\prime}) is a median triangle, this implies that J separates x and y. If z belongs to the same sector S delimited by J as x or y, then we deduce that \bigcap\mathcal{S}\subset S cannot contain x^{\prime} or y^{\prime}, which is absurd. Therefore, J separates x,y,z. To prove the converse, we will show the following more general assertion:
Fact 2.89.
Let (x^{\prime\prime},y^{\prime\prime},z^{\prime\prime}) be any median triangle of (x,y,z). If a hyperplane J separates x,y,z, then it must separate x^{\prime\prime} and y^{\prime\prime}.
According to Lemma 2.87, there exists some \mathcal{S}_{0}\subset\mathcal{S} such that x^{\prime\prime},y^{\prime\prime},z^{\prime\prime} are respectively the projections of x,y,z onto \bigcap\mathcal{S}_{0}. Since it follows from Lemma 2.85 that a hyperplane separating x and x^{\prime\prime} does not separate y and z, we know that J cannot sepate x and x^{\prime\prime}. Similarly, it cannot separate y and y^{\prime\prime}. On the other hand, we know that J separates x and y, so we deduce that J must separate x^{\prime\prime} and y^{\prime\prime}. This concludes the proof of our fact, and also of our claim.
Similarly, one shows that the hyperplanes separating x^{\prime} and z^{\prime}, or y^{\prime} and z^{\prime}, are precisely the hyperplanes separating x,y,z. Thus, we have proved that (x^{\prime},y^{\prime},z^{\prime}) is an equilateral median triangle of (x,y,z) of size the number of hyperplanes separating x,y,z, which we denote by \mu(x,y,z). On the other hand, it follows from Fact 2.89 that the size of any median triangle is at least \mu(x,y,z). Thus, we have proved that (x^{\prime},y^{\prime},z^{\prime}) is a quasimedian of (x,y,z).
We record the following assertion which we have proved for future use:
Fact 2.90.
For every triple (x,y,z), its quasimedian (x^{\prime},y^{\prime},z^{\prime}) satisfies the following property: a hyperplane separates x^{\prime} and y^{\prime} if and only if it separates x^{\prime},y^{\prime},z^{\prime} if and only if it separates x,y,z.
Now, we want to prove that it is the unique quasimedian of (x,y,z). So let (x^{\prime\prime},y^{\prime\prime},z^{\prime\prime}) be another equilateral median triangle of (x,y,z). According to Lemma 2.87, there exists some \mathcal{S}_{0}\subset\mathcal{S} such that x^{\prime\prime},y^{\prime\prime},z^{\prime\prime} are respectively the projections of x,y,z onto \bigcap\mathcal{S}_{0}. If \bigcap\mathcal{S}_{0}=\bigcap\mathcal{S}, then (x^{\prime\prime},y^{\prime\prime},z^{\prime\prime})=(x^{\prime},y^{\prime},z^% {\prime}), so suppose that \bigcap\mathcal{S}\nsubseteq\bigcap\mathcal{S}_{0}. By considering a hyperplane separating a vertex of \bigcap\mathcal{S}_{0}\backslash\bigcap\mathcal{S} from its projection onto \mathcal{S}, we find a hyperplane J intersecting \bigcap\mathcal{S}_{0} which is disjoint from \bigcap\mathcal{S}. By applying Lemma 2.85, it follows that J delimits a sector S which contains an element of S^{\prime}\in\mathcal{S}; say that y,z\in S^{\prime} and x\notin S^{\prime}. Let J^{\prime} denote the hyperplane delimiting the sector S^{\prime}. Notice that y^{\prime\prime}\in\bigcap\mathcal{S}_{0}\subset S^{\prime}, and similarly z^{\prime\prime}\in S^{\prime}. On the other hand, we claim that x^{\prime\prime}\notin S^{\prime}. Otherwise, J^{\prime} separates x and x^{\prime\prime} because x\notin S^{\prime}, and it follows from Lemma 2.34 that J^{\prime} separates x from \bigcap\mathcal{S}_{0}. Therefore, it is sufficient to show that J^{\prime} intersects \bigcap\mathcal{S}_{0}. But S^{\prime}\subset S implies that
S^{c}\cap\bigcap\mathcal{S}_{0}\subset(S^{\prime})^{c}\bigcap\mathcal{S}_{0}, 
and because J intersects \bigcap\mathcal{S}_{0}, we know that the lefthand side is non empty. Our claim follows. Thus, we have proved that J^{\prime} separates x and \{y,z\}, but does not separate y and z. On the other hand, we know from Fact 2.89 that any hyperplane separating x,y,z separates x^{\prime\prime} and y^{\prime\prime}. Therefore, d(x^{\prime\prime},y^{\prime\prime})\geq\mu(x,y,z)+1. A fortiori, the size of our equilateral median triangle must be strictly larger than the size of (x^{\prime},y^{\prime},z^{\prime}), so (x^{\prime\prime},y^{\prime\prime},z^{\prime\prime}) is not a quasimedian of (x,y,z). This proves that (x^{\prime},y^{\prime},z^{\prime}) is the unique quasimedian of (x,y,z).
Finally, let P denote the gated hull of \{x^{\prime},y^{\prime},z^{\prime}\}. As a consequence of Fact 2.89, any hyperplane separating x,y,z must separate x^{\prime},y^{\prime},z^{\prime}. Since d(x^{\prime},y^{\prime})=d(y^{\prime},z^{\prime})=d(z^{\prime},x^{\prime})=\mu% (x,y,z), we deduce that no other hyperplane can separate two vertices of \{x^{\prime},y^{\prime},z^{\prime}\}. Therefore, it follows from Proposition 2.68 that the hyperplanes of P correspond to the hyperplanes of X separating x,y,z. In particular, they must be pairwise transverse, so that Lemma 2.74 implies that P is a prism. This proves the last statement of our proposition. ∎
Remark 2.91.
It can be proved that the convex hull of the quasimedian of any triple of vertices is a product of triangles. Indeed, it follows from the description of convex hulls given in Section 2.8 that the sectors of the convex hull of some set of vertices S\subset X, viewed as a quasimedian graph on its own right, correspond to the sectors of X separating two vertices of S; as a consequence, the hyperplanes of the convex hull of our quasimedian delimits exactly three sectors, so that it follows from Lemma 2.74 that this convex hull is isomorphic to a product of triangles. This statement was originally proved by Mulder in his thesis [Mul80, (25) p. 149] (see also [BC08, Proposition 3]).
Corollary 2.92.
A graph is median if and only if it is quasimedian and trianglefree.
Proof.
Proving that a median graph is quasimedian and trianglefree is left as an exercice. Conversely, if X is a quasimedian graph which is trianglefree, then any hyperplane delimits exacly two sectors. In particular, no hyperplane can separate three vertices at the same time. Therefore, the unique quasimedian given by Proposition 2.84 has size zero, ie., it produces a unique median vertex. ∎
Next, let us prove the following application of Corollary 2.92. Recall that, given a graph X and two vertices x,y\in X, the interval I(x,y) is defined as the union of all the geodesics between x and y.
Proposition 2.93.
In a quasimedian graph, an interval is median.
We begin by proving a technical lemma.
Lemma 2.94.
Let X be a quasimedian graph, x,y\in X two vertices and a,b\in I(x,y) two adjacent vertices. Then d(x,a)\neq d(x,b).
Proof.
Suppose by contradiction that d(x,a)=d(x,b)=k. We deduce from the triangle condition that there exists some vertex p\in X adjacent to both a and b such that d(x,p)=k1. Similarly, noticing that
d(y,a)=d(x,y)d(x,a)=d(x,y)d(x,b)=d(y,b), 
we know that there exists some vertex q\in X adjacent to both a and b such that d(y,q)=d(y,a)1=d(x,y)d(x,a)1=d(x,y)k1. Now,
d(x,p)+d(y,q)+2=d(x,y)\leq d(x,p)+d(p,q)+d(q,y), 
hence d(p,q)\geq 2. On the other hand d(p,q)\leq d(p,a)+d(a,q)=2, so d(p,q)=2. As a consequence, the vertices a,b,p,q define an induced subgraph which is isomorphic to K_{4}^{}, a contradiction. ∎
The previous lemma has several interesting consequences on the structure of intervals.
Corollary 2.95.
In a quasimedian graph, an interval is an induced subgraph.
Proof.
Let X be a quasimedian graph and x,y\in X two vertices. If a,b\in I(x,y) are two vertices adjacent in X, we want to prove that they are adjacent in I(x,y). First, the previous lemma implies that d(x,a)\neq d(x,b); say d(x,a)=d(x,b)1. Let \gamma denote the concatenation of the subsegment between x and a of some geodesic between x and y passing through a, followed by the edge (a,b), and then followed by the subsegment between b and y of some geodesic between x and y passing through b. Now, notice that
d(x,y)=d(x,b)+d(b,y)=d(x,a)+1+d(b,y)=d(x,a)+d(a,b)+d(b,y), 
which is precisely the length of \gamma. Therefore, \gamma defines a geodesic between x and y containing the edge (a,b). A fortiori, (a,b)\subset I(x,y). ∎
Corollary 2.96.
In a quasimedian graph, an interval is trianglefree.
Proof.
Let X be a quasimedian graph, x,y\in X two vertices and a,b,c\in I(x,y) three vertices such that d(x,b)\leq d(x,a),d(x,c) and b is adjacent to both a and c. Because a and b are adjacent, and that d(x,b)\leq d(x,a), we deduce that either d(x,b)=d(x,a) or d(x,b)=d(x,a)1; but the former case is impossible according the previous lemma, hence d(x,b)=d(x,a)1. Similarly, we show that d(x,c)=d(x,b)+1, so that d(x,c)=d(x,a). It follows from the previous lemma that a and c cannot be adjacent, so that the vertices a,b,c cannot define a triangle. ∎
The following lemma is the last step to prove Proposition 2.93.
Lemma 2.97.
In a quasimedian graph, an interval is a convex subgraph.
Proof.
Let X be a quasimedian graph and x,y\in X two vertices. We want to prove that, if c is a vertex which belongs to some geodesic [a,b] between two vertices a,b\in I(x,y), then c\in I(x,y), ie., there exists some geodesic between x and y passing through c. More precisely, if we fix two geodesics [x,c] and [c,y], we claim that the concatenation [x,c]\cup[c,y] defines a geodesic. For convenience, we will denote by [x,a] (resp. [a,y]) the subsegment between x and a (resp. a and y) of some geodesic between x and y passing through a; and similarly for [x,b] and [b,y].
Suppose by contradiction that [x,c]\cup[c,y] is not a geodesic, so that there exists some hyperplane J intersecting both [x,c] and [c,y]. We will use the following observation:
Fact 2.98.
Let [p,q]\cup[q,r]\cup[r,p] be a geodesic triangle and J a hyperplane intersecting [p,q]. Then J has to intersect either [q,r] or [r,p].
Indeed, because J intersects the geodesic [p,q], necessarily J separates [p,q]. Therefore, if the path [q,r]\cup[r,p] does not meet J, it defines a path between p and q which lies in some sector delimited by J, which is impossible. Consequently, J has to intersect either [q,r] or [r,p] (or both), proving our fact.
Let [a,c] (resp. [b,c]) denote the subsegment of [a,b] between a and c (resp. b and c). By applying our fact to the triangle [a,x]\cup[x,c]\cup[c,a], we deduce that J has to intersect either [x,a] or [a,c]; and by applying our fact to the triangle [a,y]\cup[y,c]\cup[c,a], we deduce that J has to intersect either [y,a] or [a,c]. On the other hand, because [x,a]\cup[a,y] is a geodesic, J intersects it at most once. Therefore, J must intersect [a,c]. Similarly, we show that J intersects [b,c]. However, [a,c]\cup[c,b]=[a,b] is a geodesic, so J cannot intersect it twice, hence a contradiction. ∎
Proof of Proposition 2.93..
Let X be a quasimedian graph and x,y\in X two vertices. Because I(x,y) is a convex subgraph of X, the interval I(x,y) is a quasimedian graph on its own right. Moreover, we know thanks to Corollary 2.96 that I(x,y) is trianglefree, so we conclude from Corollary 2.92 that I(x,y) is a median graph. ∎
Remark 2.99.
The uniqueness of quasimedians stated in Proposition 2.84 was proved in [BMW94], as well as the gatedness of intervals and Proposition 2.93; and Corollary 2.92 was proved in [Mul80]. It is interesting to notice that our langage of hyperplanes turns out to define a common framework to prove all these results.
2.8 Convex hulls
Definition 2.100.
Let X be a graph and S\subset X a set of vertices. The convex hull of S is the smallest convex subgraph of X containing S; alternatively, this is the intersection of all the convex subgraphs of X which contain S.
In this section, we are interested in characterizing the convex hull of a given set of vertices. In CAT(0) cube complexes, the convex hull of some subspace coincides with the intersection of all the halfspaces containing it. However, according to our dictionnary given by Table 1, the translation of this statement is precisely Lemma 2.69. So we need to introduce another family of specific subgraphs.
Definition 2.101.
Let X be a quasimedian graph. A multisector M is the subgraph generated by a union D_{1}\cup\cdots\cup D_{n}, where D_{1},\ldots,D_{n} are sectors delimited by a common hyperplane. In particular, if D is a sector delimited by a hyperplane J, we refer to the multisector generated by the sectors delimited by J which are different from D as the cosector associated to D.
As a consequence of the following lemma, a proper multisector is never gated.
Lemma 2.102.
Let J be a hyperplane and D_{1},\ldots,D_{k} some sectors delimited by J, with k\geq 2. Let M denote the multisector defined by D_{1},\ldots,D_{k}. The gated hull of M is N(J)\cup D_{1}\cup\cdots\cup D_{k}.
Proof.
Let G denote the gated hull of M, so that we want to prove that
G=N(J)\cup D_{1}\cup\cdots\cup D_{k}. 
Let C be a clique dual to J. Because k\geq 2, M contains an edge of C, so that G must contain C since any gated subgraph contains its triangles. Therefore, N(J)\cup D_{1}\cup\cdots\cup D_{k}\subset G. To conclude, it is sufficient to notice that N(J)\cup D_{1}\cup\cdots\cup D_{k} is gated. This is proved by the following fact, which is slightly more general than what we need here since we allow k=1.
Fact 2.103.
Let J be a hyperplane and D_{1},\ldots,D_{k} some sectors delimited by J with k\geq 1. Then N(J)\cup D_{1}\cup\cdots\cup D_{k} is a gated subgraph.
For convenience, set L=N(J)\cup D_{1}\cup\cdots\cup D_{k}. Let x\in X\backslash L be a vertex. Because J separates x from D_{1},\ldots,D_{k}, any geodesic between x and some vertex of D_{1}\cup\cdots\cup D_{k} must intersect N(J). Therefore, d(x,L)=d(x,N(J)). As a consequence, the gate of x in N(J) defines a gate of x in L. We deduce that L must be gated. ∎
The main result of this section is the following statement.
Proposition 2.104.
The convex hull of a set of vertices S coincides with the intersection of all the multisectors containing S
Our proposition will be essentially a consequence of the following two preliminary lemmas.
Lemma 2.105.
A multisector is convex.
Proof.
Let M be a multisector generated by some sectors D_{1},\ldots,D_{k} delimited by a hyperplane J. If x,y\in M are two vertices, we want to prove that any geodesic \gamma between x and y belongs to M. Let 1\leq i,j\leq k be such that x\in D_{i} and y\in D_{j}. If i=j, then \gamma\subset D_{i}\subset M by convexity of D_{i}, and we are done. Now, suppose that i\neq j. In particular, J separates x and y, so that \gamma must contain an edge e dual to J. Let us write \gamma as the concatenation \gamma_{1}\cup e\cup\gamma_{2}, where \gamma_{1} is the initial segment of \gamma whose last vertex is the initial point of e and \gamma_{2} the final segment of \gamma starting from the terminating vertex of e. We know that, because \gamma is a geodesic, J intersects it once, hence \gamma_{1}\subset D_{i} and \gamma_{2}\subset D_{j}; in particular, \gamma_{1},\gamma_{2}\subset M. Moreover, because the endpoints of e belong to M, we necessarily have e\subset M. A fortiori, we have \gamma\subset M, concluding the proof. ∎
Lemma 2.106.
Let C be a convex subgraph and x\notin C a vertex. There exists a sector D satisfying x\in D and D\cap C=\emptyset.
Proof.
Let y\in C be a vertex minimizing the distance ot x in C. Let J be a hyperplane separating x and y (such a hyperplane exists since x\notin C), and let D denote the sector delimited by J which contains x. We claim that D is the sector we are looking for, ie., D\cap C=\emptyset. Fix some vertex z\in C, and let (x^{\prime},y^{\prime},z^{\prime}) be the quasimedian of the triple (x,y,z). Fix some geodesics [z,z^{\prime}], [y^{\prime},z^{\prime}], [y,y^{\prime}]. Notice that, because C is convex, the geodesic [y,y^{\prime}]\cup[y^{\prime},z^{\prime}]\cup[z^{\prime},z] between y and z must be included into C. As a consequence, because y minimizes the distance to x in C, necessarily y=y^{\prime}. In particular, the hyperplane J, which separates x and y, either separates x and x^{\prime} or it separates x^{\prime} and y^{\prime}. In the former case, J separates x and z, hence z\notin D; in the latter case, we deduce from Fact 2.90 that J separates x,y,z, hence z\notin D. We conclude that D\cap C=\emptyset. ∎
Proof of 2.104..
Let C denote the convex hull of S and I the intersection of all the multisectors containing S. We deduce from Lemma 2.105 that I is convex, hence C\subset I. Now, if x\notin C, then we deduce from Lemma 2.106 that there exists some sector D satisfying x\in D and D\cap C=\emptyset. In particular, if M denotes the cosector associated to D, then S\subset C\subset M and x\notin M. A fortiori, x\notin I since I\subset M. Thus, we have proved that X\backslash C\subset X\backslash I, or equivalently I\subset C. We conclude that I=C. ∎
Corollary 2.107.
The convex hull of a finite subset is finite.
Proof.
Let C denote the convex hull of a finite set of vertices S, and let D be a sector. Three cases may happen. Either D\cap S=\emptyset, so that S is included into the cosector associated to D, hence D\cap C=\emptyset; or S\subset D, so that C\subset D; or D separates two vertices of S. Therefore, a sector separates two vertices of C if and only if it separates two vertices of S. On the other hand, because S is finite, there exist only finitely many such sectors, so that we deduce from Corollary 2.65 that C is finite. ∎
2.9 Flat rectangles
In this section, we introduce a particular class of subgraphs we call flat rectangles. We used them in [Gen16b] to study hyperbolicity in CAT(0) cube complexes. Our goal here is to prove that similar results still hold in quasimedian graphs. In particular, Proposition 2.111 and Proposition 2.113 for CAT(0) cube complexes correspond respectively to [Gen16b, Theorem 2.13 and Theorem 3.3]. The hyperbolicity of quasimedian graphs was also studied in [CDE{}^{+}, Corollary 5].
Definition 2.108.
A flat rectangle is an isometric embedding R:[0,n]\times[0,m]\hookrightarrow X, where we identify the square complex [0,n]\times[0,m] with its 1skeleton; if m=n, R will be referred to as a flat square. If n,m\leq C for some C\geq 0, we say that R is Cthin.
The following lemma provides a useful way to construct flat rectangles.
Definition 2.109.
A quadrangle is a quadruple (a,b,c,d) satisfying b,d\in I(a,c) and a,c\in I(b,d).
Lemma 2.110.
Let (a,b,c,d) be a quadrangle and [a,b],[b,c] two geodesics. There exists a flat rectangle [0,r]\times[0,s]\hookrightarrow X satisyfing [0,r]\times\{0\}=[a,b], \{r\}\times[0,s]=[b,c] and (0,s)=d.
Proof.
According to Proposition 2.63, we can identify X with the quasicubulation of its canonical structure of space with partitions. In particular, the geodesic [a,b] produces the sequence of vertices
\sigma_{a},\ [\sigma_{a},A_{1}],\ [\sigma_{a},A_{1},A_{2}],\ldots,[\sigma_{a},% A_{1},\ldots,A_{p}]=\sigma_{b}; 
and similarly, the geodesic [b,c] produces
\sigma_{b},\ [\sigma_{b},B_{1}],\ [\sigma_{b},B_{1},B_{2}],\ldots,[\sigma_{b},% B_{1},\ldots,B_{q}]=\sigma_{c}. 
Let J_{1},\ldots,J_{p} (resp. H_{1},\ldots,H_{q}) denote the hyperplanes underlying the sectors A_{1},\ldots,A_{p} (resp. B_{1},\ldots,B_{q}). We claim that, for every 1\leq i\leq p and 1\leq j\leq q, the hyperplanes J_{i} and H_{j} are transverse. For convenience, fix two geodesics [a,d] and [d,c]; notice that, because d\in I(a,c), the concatenation [a,d]\cup[d,c] is a geodesic.
Because J_{i} intersects a geodesic between a and c, namely [a,b]\cup[b,c], necessarily J_{i} separates a and c. In particular, J_{i} must intersect the geodesic [a,d]\cup[d,c]. On the other hand, since a\in I(b,d), the concatenation [b,a]\cup[a,d] is a geodesic, so that J_{i} cannot intersect it twice. We deduce that J_{i} intersects [c,d]. Therefore, b,c\in A_{i} and a,d\notin A_{i}. Similarly, we show that a,b\in B_{j} and c,d\notin B_{j}. In particular, J_{i} and H_{j} must be transverse.
As a consequence of Lemma 2.49, for every 1\leq i\leq p and 1\leq j\leq q we have
c=[\sigma_{a},A_{1},\ldots,A_{p},B_{1},\ldots,B_{q}]=[\sigma_{a},A_{1}\ldots,A% _{i},B_{1},\ldots,B_{q},A_{i+1},\ldots,A_{p}], 
and \sigma_{ij}=[\sigma,A_{1},\ldots,A_{i},B_{1},\ldots,B_{j}] defines an orientation, ie., a vertex of X. Consider the application
R:\left\{\begin{array}[]{ccc}[0,p]\times[0,q]&\longrightarrow&X\\ (i,j)&\longmapsto&\sigma_{ij}\end{array}\right. 
We claim that R is the flat rectangle we are looking for. To conclude, we only have to prove that R is an isometric embedding. Once again according to Lemma 2.49, we know that
\sigma_{i+r,j+s}=[\sigma_{ij},A_{i+1},\ldots,A_{i+r},B_{j+1},\ldots,B_{j+s}]. 
Moreover, we deduce from Lemma 2.50 that
A_{i+k}\neq\sigma_{a}(J_{k})=\sigma_{ij}(J_{k})\ \text{and}\ B_{j+h}\neq\sigma% _{b}(H_{k})=\sigma_{a}(H_{k})=\sigma_{ij}(H_{k}). 
Therefore, according to Corollary 2.51, we conclude that d(\sigma_{ij},\sigma_{i+r,j+s})=r+s. ∎
Now, we are ready to prove the following proposition, which was fundamental in [Gen16b]. A sequence of subgraphs (Y_{1},\ldots,Y_{n}) is called a cycle of subgraphs if Y_{i}\cap Y_{i+1}\neq\emptyset for every i mod n.
Proposition 2.111.
Let X be a quasimedian graph and (Y_{1},Y_{2},Y_{3},Y_{4}) a cycle of gated subgraphs. There exists a flat rectangle [0,a]\times[0,b]\hookrightarrow X satisyfing [0,a]\times\{0\}\subset Y_{1}, [0,a]\times\{b\}\subset Y_{3}, \{a\}\times[0,b]\subset Y_{2} and \{0\}\times[0,b]\subset Y_{4}.
Proof.
Let a\in Y_{1}\cap Y_{2} be a vertex minimizing the distance to Y_{3}\cap Y_{4}. Let b (resp. d,c) denote its projection onto Y_{3} (resp. Y_{4}, Y_{3}\cap Y_{4}). It is worth noticing that, according to Lemma 2.39, we have b\in Y_{2} and d\in Y_{1}. We claim that (a,b,c,d) defines a quandrangle. For convenience, fix four geodesics [a,b], [b,c], [c,d] and [d,a].
First, c\in Y_{3} implies b\in I(a,c), and c\in Y_{4} implies d\in I(a,c).
Now, notice that, if J is a hyperplane intersecting [c,d], because c is the projection of d onto Y_{3} as a consequence of Lemma 2.38, we deduce from Lemma 2.34 that J has to be disjoint from Y_{3}. In particular, it cannot intersect [b,c]. Similarly, we show that no hyperplane intersecting [b,c] can intersect [c,d]. Therefore, the concatenation [d,c]\cup[c,b] is a geodesic according to Proposition 2.30. As a consequence, c\in I(b,d).
Finally, we want to prove that a\in I(b,d). Because we know from Proposition 2.93 that I(a,c) is a median graph, the triple (a,b,d) admits a median vertex m. Noticing that m\in I(a,b)\subset Y_{2} since Y_{2} is convex, and similarly m\in I(a,d)\subset Y_{1} since Y_{1} is convex, we deduce that m\in Y_{1}\cap Y_{2}. On the other hand,
d(a,c)=d(a,d)+d(d,c)=d(a,m)+d(m,d)+d(d,c)=d(a,m)+d(m,c), 
hence d(m,c)=d(a,c)d(a,m). Since a minimizes the distance to Y_{3}\cap Y_{4} in Y_{1}\cap Y_{2}, we deduce that a=m, hence a=m\in I(b,d).
Thus, we have proved that (a,b,c,d) defines a quadrangle. According to Lemma 2.110, there exists a flat rectangle [0,n]\times[0,m]\hookrightarrow X such that (0,0)=a, (n,0)=b, (n,m)=c and (0,m)=d. Because a,b\in Y_{2} and Y_{2} is convex, necessarily [0,n]\times\{0\}\subset Y_{2}. Similarly, we show that \{n\}\times[0,m]\subset Y_{3}, [0,n]\times\{m\}\subset Y_{4} and \{0\}\times[0,m]\subset Y_{1}. ∎
Corollary 2.112.
Let Y_{1},Y_{2}\subset X be two gated subgraphs. If d denote the diameter of the projection of Y_{1} onto Y_{2}, then there exists a flat rectangle [0,d]\times[0,p]\hookrightarrow X satisfying [0,d]\times\{0\}\subset Y_{1} and [0,d]\times\{p\}\subset Y_{2}.
Proof.
Let x,y\in Y_{1} be two vertices such that, if x^{\prime},y^{\prime} denote their respective projections onto Y_{2}, then d(x^{\prime},y^{\prime})=d. Let J_{1},\ldots,J_{d} denote the d hyperplanes separating x^{\prime} and y^{\prime}. For every 1\leq i\leq d, we denote by J_{i}^{+} (resp. J_{i}^{}) the sector delimited by J_{i} which contains x^{\prime} (resp. y^{\prime}). Finally, let C^{\pm}=\bigcap\limits_{i=1}^{d}J_{i}^{\pm}. Notice that x^{\prime}\in C^{+} and y^{\prime}\in C^{}, so that Y_{2}\cap C^{+} and Y_{2}\cap C^{} are non empty. Then, as a consequence of Lemma 2.34, any hyperplane separating x and x^{\prime} must be disjoint from Y_{2}, so that no J_{i} separates x and x^{\prime}, hence x\in J_{i}^{+} and finally x\in C^{+}. Similarly, we show that y\in C^{}. Therefore, the intersection Y_{1}\cap C^{+} and Y_{1}\cap C^{} are non empty. We have prove that (Y_{1},C^{+},Y_{2},C^{}) is a cycle of gated subgraphs, so that Proposition 2.111 implies that there exists a flat rectangle [0,n]\times[0,m]\hookrightarrow X satisfying [0,n]\times\{0\}\subset Y_{1}, [0,n]\times\{m\}\subset Y_{2}, (0,m)\in C^{+} and (n,m)\in C^{}. Since J_{1},\ldots,J_{d} clearly separate C^{+} and C^{}, we have m\geq d, so that taking a subrectangle of [0,n]\times[0,m] if necessary produces the flat rectangle we are looking for. ∎
As an application, we are able to characterize hyperbolic quasimedian graphs. First, define a facing triple as the data of three hyperplanes such that no one separates the two others, and a join of hyperplanes (\mathcal{H},\mathcal{V}) as the data of two collections of hyperplanes \mathcal{H},\mathcal{V} which do not contain facing triples and such that any hyperplane of \mathcal{H} is transverse to any hyperplane of \mathcal{V}; a join (\mathcal{H},\mathcal{V}) is Kthin, where K\geq 0 is some fixed integer, if \min(\#\mathcal{H},\#\mathcal{V})\leq K.
Proposition 2.113.
Let X be a quasimedian graph. The following assertions are equivalent:

(i)
X is hyperbolic;

(ii)
its join of hyperplanes are uniformly thin;

(iii)
its flat rectangles are uniformly thin;

(iv)
its bigons are uniformly thin.
Let us begin by proving a preliminary lemma.
Lemma 2.114.
Let X be a quasimedian graph and x,y\in X two vertices. For every vertex z\in I(x,y) and for every geodesic [x,y] between x and y, there exists a flat square R:[0,n]\times[0,n]\hookrightarrow X satisfying z=(0,0) and (n,n)\in[x,y].
Proof.
Let w be the unique vertex of [x,y] satisfying d(x,w)=d(x,z). Because I(x,y) is median according to Proposition 2.93, we can introduce the median vertex a\in X of the triple (w,x,z) and the median vertex b\in X of the triple (w,y,z). Notice that a,b\in I(z,w). Moreover, if we fix some geodesics [x,a], [a,z], [z,b] and [b,y], we notice that the concatenation
[x,a]\cup[a,z]\cup[z,b]\cup[b,y] 
defines a geodesic; in particular, [a,z]\cup[z,b] is a geodesic between a and b passing through z, hence z\in I(a,b). Similarly, we show that w\in I(a,b). Therefore, there exists a flat rectangle R with a,b,w,z as corners. By proving that
d(a,z)=d(a,w)=d(b,z)=d(b,w), 
we will justify that this flat rectangle is the flat square we are looking for. First, notice that
d(x,a)+d(a,z)=d(x,z)=d(x,w)=d(x,a)+d(a,w), 
hence d(a,z)=d(a,w). Let \ell_{1} denote this common value. Similarly, because
d(y,z)=d(x,y)d(x,z)=d(x,y)d(x,w)=d(y,w), 
we show that d(b,z)=d(b,w). Let \ell_{2} denote this common value. Next, notice that
2\ell_{1}=d(z,a)+d(a,w)=d(z,b)+d(b,w)=2\ell_{2}, 
hence \ell_{1}=\ell_{2}. This concludes the proof. ∎
Proof of Proposition 2.113..
Suppose that X is \deltahyperbolic for some \delta\geq 0 and let (\mathcal{H},\mathcal{V}) be a join of hyperplanes. First notice that the cubical dimension of X must be finite. If it was not the case, X would contain (the 1skeleton of) an ncube as an isometrically embedded subgraph for every n\geq 1 according to Proposition 2.73 and Lemma 2.80, but such a cube contains a triangle which is not (n1)thin. Next, suppose that \#\mathcal{H},\#\mathcal{V}\geq\mathrm{Ram}(d) for some d\geq\dim_{\square}(X), where \mathrm{Ram}(\cdot) denotes the corresponding Ramsey number (which is defined by the following property: for every r\geq 1 and every edgecoloring of the complete graph on at least \mathrm{Ram}(r) vertices with two colors, one can find a monochromatic complete subgraph with at least r vertices). This implies that \mathcal{H} and \mathcal{V} contain respectively some subcollections \mathcal{H}_{0} and \mathcal{V}_{0} of d pairwise non transverse hyperplanes (this is a classical argument; see for instance [Gen16b, Lemma 3.7]). Because \mathcal{H}_{0} contains no facing triple, we can number its elements
\mathcal{H}_{0}=\{H_{1},\ldots,H_{d}\} 
so that, for every 2\leq i\leq d1, the hyperplane H_{i} separates H_{i1} and H_{i+1}. Number similarly the elements of \mathcal{V}_{0}:
\mathcal{V}_{0}=\{V_{1},\ldots,V_{d}\}, 
i.e., V_{i} separates V_{i1} and V_{i+1} for every 2\leq i\leq d1. Now, (N(V_{1}),N(H_{1}),N(V_{d}),N(H_{d})) defines a cycle of gated subgraphs, so Proposition 2.111 yields a flat rectangle R:[0,n]\times[0,m]\hookrightarrow X such that [0,n]\times\{0\}\subset N(H_{1}), \{n\}\times[0,m]\subset N(V_{d}), [0,n]\times\{m\}\subset N(H_{d}) and \{0\}\times[0,m]\subset N(V_{1}). Since V_{1} and V_{d} are separated by d2 hyperplanes, it follows that n\geq d2; similarly, we know that m\geq d2. On the other hand, because a flat square [0,k]\times[0,k]\hookrightarrow X contains a geodesic triangle which is not (k1)thin, we deduce that the flat rectangles of X are all (\delta+1)thin. Therefore, d\leq\min(n,m)+2\leq\delta+3. Consequently, we have proved that
\min(\#\mathcal{H},\#\mathcal{V})\leq\max\left(\mathrm{Ram}(\dim_{\square}(X))% ,\mathrm{Ram}(\delta+4)\right). 
This proves the implication (i)\Rightarrow(ii).
Next, because the hyperplanes dual to a flat rectangle R:[0,n]\times[0,m]\hookrightarrow X defines a join of hyperplanes (\mathcal{H},\mathcal{V}) satisfying \#\mathcal{H}=m and \#\mathcal{V}=n, we know that the implication (ii)\Rightarrow(iii) holds.
Now, suppose that (iii) holds, ie., there exists some constant B such that the flat rectangles of X are all Bthin. So, if we fix some bigon (\gamma_{1},\gamma_{2}) and some vertex x\in\gamma_{1}, Lemma 2.114 implies that there exists a flat square R such that x\in R and R\cap\gamma_{2} contains some vertex y, and we conclude that
d(x,\gamma_{2})\leq d(x,y)\leq\mathrm{diam}(R)\leq 2B. 
This proves the implication (iii)\Rightarrow(iv). Finally, suppose that (iv) holds, ie., there exists some constant B such that any bigon of X is Bthin. Let [x,y]\cup[y,z]\cup[z,x] be a geodesic triangle and let (x^{\prime},y^{\prime},z^{\prime}) denote the quasimedian of the triple (x,y,z). We fix some geodesics [x,x^{\prime}], [y,y^{\prime}], [z,z^{\prime}], [x^{\prime},y^{\prime}], [y^{\prime},z^{\prime}], [z^{\prime},x^{\prime}]. We want to prove that, if p\in[x,y] is a vertex, then d(p,[y,z]\cup[z,x])\leq 4B+3. Because the bigon defined by [x,y] and [x,x^{\prime}]\cup[x^{\prime},y^{\prime}]\cup[y^{\prime},y] is Bthin by assumption, there exists some q\in[x,x^{\prime}]\cup[x^{\prime},y^{\prime}]\cup[y^{\prime},y] such that d(p,q)\leq B. If q\in[x,x^{\prime}] (resp. q\in[y,y^{\prime}]), then similarly we find some q^{\prime}\in[x,z] (resp. q^{\prime}\in[y,z]) such that d(q,q^{\prime})\leq B, hence
d(p,[x,x^{\prime}]\cup[x^{\prime},y^{\prime}]\cup[y^{\prime},y])\leq d(p,q^{% \prime})\leq d(p,q)+d(q,q^{\prime})\leq 2B. 
From now on, suppose that q\in[x^{\prime},y^{\prime}]. Fix some vertex q^{\prime}\in[x^{\prime},z^{\prime}]. Again by the same argument, we find some vertex q^{\prime\prime}\in[x,z] satisfying d(q^{\prime},q^{\prime\prime})\leq B. According to Proposition 2.84, the gated hull of \{x^{\prime},y^{\prime},z^{\prime}\} is a prism P, so
d(p,[x,x^{\prime}]\cup[x^{\prime},y^{\prime}]\cup[y^{\prime},y])\leq d(p,q^{% \prime\prime})\leq d(p,q)+d(q,q^{\prime})+d(q^{\prime},q^{\prime\prime})\leq 2% B+\mathrm{diam}(P). 
On the other hand, a prism of cubical dimension 2n contains a bigon which is not (n1)thin, hence \dim_{\square}(P)\leq 2B+1; as a conquence, \mathrm{diam}(P)\leq 2B+3. Therefore,
d(p,[x,x^{\prime}]\cup[x^{\prime},y^{\prime}]\cup[y^{\prime},y])\leq 2B+2B+3=4% B+3. 
Thus, we have proved that the geodesic triangles of X are all (4B+3)thin, so that X must be hyperbolic. This proves (iv)\Rightarrow(i). ∎
2.10 Fixed point theorem
This section is dedicated to the proof of the following result.
Theorem 2.115.
A group acting on a quasimedian graph with a bounded orbit stabilises a finite prism.
In the particular case of a finite group acting on a locally finite quasimedian graph, the previous statement follows from [BCC{}^{+}13a, Theorem 4]. Our proof follows the argument given by Roller in [Rol98, Theorem 11.7], where he shows a similar statement for median algebras (which holds in particular for CAT(0) cube complexes). We begin by proving our theorem for finite quasimedian graphs.
Proposition 2.116.
Let X be a finite quasimedian graph. There exists a prism P\subset X which is stabilised by \mathrm{Aut}(X).
Proof.
Let \mathfrak{D} denote the sectors D of X satisfying D>\frac{1}{2}X (by abuse of notation, we denote by \cdot the number of vertices of the subgraph which we consider), and let X^{(1)} denote the (finite) intersection \bigcap\limits_{D\in\mathfrak{D}}D. Notice that, if D_{1},D_{2}\in\mathfrak{D}, then D_{1}\cap D_{2}\neq\emptyset, since otherwise we would have
X\geqD_{1}+D_{2}>\frac{1}{2}X+\frac{1}{2}X=X, 
a contradiction. Therefore, according to Helly’s property 2.8, X^{(1)} is a non empty gated subgraph of X, which is clearly \mathrm{Aut}(X)invariant. If X^{(1)} is a single vertex, we are done. Otherwise, as a quasimedian graph on its own right, X^{(1)} must contain some hyperplanes. In particular, if there exists a sector D of X^{(1)} satisfying D>\frac{1}{2}X^{(1)}, then the previous construction can be iterated, and we obtain a new non empty \mathrm{Aut}(X)invariant gated subgraph X^{(2)}, and so on. Because X is finite, our sequence X\supsetneq X^{(1)}\supsetneq X^{(2)}\supsetneq\cdots must terminate to some non empty \mathrm{Aut}(X)invariant gated subgraph Y. By construction, either Y is a single vertex, so that we are done, or any sector D of Y satisfies D\leq\frac{1}{2}Y.
In the latter case, we claim that any two hyperplanes of Y are transverse. If not, there exist two nested hyperplanes \mathcal{P} and \mathcal{Q}. Let P denote the sector delimited by \mathcal{P} containing \mathcal{Q}, and P^{\prime} the union of the sectors delimited by \mathcal{P} which are different from P; similarly, let Q denote the sector delimited by \mathcal{Q} containing \mathcal{P}, and Q^{\prime} the union of the sectors delimited by \mathcal{Q} which are different from Q. Notice that P^{\prime}\subset Q implies P^{\prime}\leqQ\leq\frac{1}{2}Y, hence
\frac{1}{2}Y\geqP=YP^{\prime}\geqY\frac{1}{2}Y=\frac{1}{2}Y. 
Therefore, P=\frac{1}{2}Y, and finally P^{\prime}=YP=\frac{1}{2}Y. Similarly, we show that Q^{\prime}=\frac{1}{2}Y. As a consequence,
Y\backslash(P^{\prime}\cup Q^{\prime})=YP^{\prime}Q^{\prime}=Y% \frac{1}{2}Y\frac{1}{2}Y=0, 
which is impossible. Thus, we have proved that the hyperplanes of Y are pairwise transverse. It follows from Lemma 2.74 that Y must be a prism. ∎
Definition 2.117.
Let X be a quasimedian graph. We define the gated topology on X^{(0)} by taking the set of the sectors of X as a prebasis.
Notice that a sector is open by definition, so that, because two distinct vertices are always separated by at least one hyperplane, the topology is Hausdorff. But a sector is also closed, since it is the complement of the union of the other sectors delimited by the corresponding hyperplane. In particular, the gated topology is totally disconnected and any gated subgraph is closed.
Lemma 2.118.
With respect to the gated topology, a ball is compact.
In the sequel, we will use the following notation: if S is a set and \mathcal{U} a collection of subsets, the intersection \bigcap\mathcal{U} denotes \bigcap\limits_{U\in\mathcal{U}}U.
Proof of Lemma 2.118..
Let B be a ball of radius r. Recall that a Hausdorff topological space is compact if and only if, for every collection \mathcal{F} of closed subspaces satisfying \bigcap\mathcal{F}_{s}\neq\emptyset for every finite subcollection \mathcal{F}_{s}\subset\mathcal{F}, we have \bigcap\mathcal{F}\neq\emptyset. Therefore, it is sufficient to prove that, given a collection \mathcal{H} of sectors satisfying \bigcap\mathcal{H}_{s}\cap B\neq\emptyset for every finite subcollection \mathcal{H}_{s}\subset\mathcal{H}, we have \bigcap\mathcal{H}\cap B\neq\emptyset.
If y\in B, let \mathcal{H}_{y} denote \{H\in\mathcal{H}\mid y\notin H\}. We claim that \#\mathcal{H}_{y}\leq 2r. Let H_{1},\ldots,H_{n}\in\mathcal{H}_{y}. By our hypothesis, there exists a vertex z\in B\cap H_{1}\cap\cdots\cap H_{n}. Clearly, H_{1},\ldots,H_{n} separate y and z, hence \#\mathcal{H}_{y}\leq d(y,z)\leq 2r.
Next, we want to prove that \bigcap\mathcal{H}\neq\emptyset. Let x\in B be the center of B. Since our previous claim states that \mathcal{H}_{x} is finite, we know that there exists a vertex y\in B\cap\bigcap\mathcal{H}_{x}; notice that any sector of \mathcal{H} contains either x or y. Let \mathcal{H}_{0}=\mathcal{H}\backslash(\mathcal{H}_{x}\cup\mathcal{H}_{y}) denote the subcollection of the sectors of \mathcal{H} containing both x and y. Notice that x\in\bigcap\mathcal{H}_{0}\cap\bigcap\mathcal{H}_{y} and y\in\bigcap\mathcal{H}_{0}\cap\bigcap\mathcal{H}_{x}; moreover, because \mathcal{H}_{x} and \mathcal{H}_{y} are finite, our hypothesis implies that B\cap\bigcap\mathcal{H}_{x}\cap\bigcap\mathcal{H}_{y} is non empty, so that in particular \bigcap\mathcal{H}_{x}\cap\bigcap\mathcal{H}_{y} is necessarily non empty. Applying Helly’s property 2.8, we deduce that
\bigcap\mathcal{H}=\bigcap\mathcal{H}_{0}\cap\bigcap\mathcal{H}_{x}\cap\bigcap% \mathcal{H}_{y}\neq\emptyset. 
Let z\in\bigcap\mathcal{H}, and let (x^{\prime},y^{\prime},z^{\prime}) denote the quasimedian of the triple (x,y,z). Notice that any sector H of \mathcal{H} contains z and either x or y; a fortiori, it must contain either I(z,x) or I(z,y), hence z^{\prime}\in H. Moreover
d(x,z^{\prime})=d(x,x^{\prime})+d(x^{\prime},z^{\prime})=d(x,x^{\prime})+d(x^{% \prime},y^{\prime})=d(x,y^{\prime})\leq d(x,y)\leq r, 
hence z^{\prime}\in B. Thus, the intersection \bigcap\mathcal{H}\cap B is non empty. ∎
Lemma 2.119.
With respect to the gated topology, the set of accumulation points B(x,r)^{\prime} of a ball B(x,r) centered at x of radius r is included into B(x,r1).
Proof.
Because B(x,r) is compact, we know that B(x,r)^{\prime}\subset B(x,r). Therefore, we only have to prove that a vertex y\in X satisfying d(x,y)=r cannot belong to B(x,r)^{\prime}.
Let S denote the intersection of all the sectors of X which contain y but not x. Notice that, because there exist only finitely many sectors separating x and y, S is the intersection of finitely many sectors, so that it must be open with respect to the gated topology. Fix some vertex z\in B(x,r)\cap S. Notice that no hyperplane separate x,y,z at the same time: otherwise, there would exist a sector containing y but not x and z, contradicting our choice of z. Therefore, we deduce from Proposition 2.84 that the triple \{x,y,z\} admits a median vertex m. Similarly, no hyperplane can separate m and y, since otherwise there would exist a sector containing y but not x and z, hence m=y. In particular, y belongs to a geodesic between x and z, hence
d(x,z)=d(x,y)+d(y,z)=r+d(y,z). 
On the other hand, z\in B(x,r) implies d(x,z)\leq r, hence y=z. Thus, we have proved that B(x,r)\cap S=\{y\}. As a consequence, \{y\} is an open subspace of B(x,r), which implies that y\notin B(x,r)^{\prime}. ∎
Proof of Theorem 2.115.
Let G be a group acting on a quasimedian graph X with a bounded orbit. Let B\subset X be a bounded Ginvariant subset, say a bounded orbit of G. In particular, there exist some vertex x\in X and some integer r\geq 0 such that B\subset B(x,r). We define the sequence of subsets (B^{(i)}) by B^{(0)}=B and B^{(i+1)}=\left(B^{(i)}\right)^{\prime} for every i\geq 0. Notice that B^{\prime}\subset B(x,r)^{\prime}\subset B(x,r1) according to Lemma 2.119, and we deduce by induction that B^{(i)}\subset B(x,ri) for every i\geq 0. As a consequence, there exists some i\geq 0 such that B^{(i+1)}=\emptyset. This implies that B^{(i)} must be finite, since otherwise B^{(i+1)} would be non empty by compactness. Thus, B^{(i)} is a finite Ginvariant subset. Let Y denote its convex hull. Clearly, Y is Ginvariant, and we deduce from Corollary 2.107 that Y is finite as well. Because Y is a quasimedian graph on its own right, we deduce from Proposition 2.116 that Y contains a Ginvariant prism, which concludes the proof. ∎
2.11 CAT(0)ness
In this section, we prove that quasimedian graphs can be endowed naturally with a cellular structure making them CAT(0) spaces. See [Lea13, Appendix] and references therein for the link between CAT(0) cube complexes and CAT(0) spaces; and [BCC{}^{+}13b] to see how CAT(0) spaces arise similarly from the more general class of (finite) retracts of products of chordal graphs.
Given a complete graph K, consider the Hilbert space \mathcal{H}(K) of the \ell^{2}functions K^{(0)}\to[0,1]. If \delta_{v}, for v\in K^{(0)}, denotes the function u\mapsto\left\{\begin{array}[]{cl}0&\text{if}\ u\neq v\\ 1&\text{otherwise}\end{array}\right., then the map v\mapsto\frac{1}{\sqrt{2}}\delta_{v} defines an isometric embedding K\hookrightarrow\mathcal{H}(K). By identifying K with the convex hull of its image in \mathcal{H}(X), we naturally view K as a CAT(0) cellular complex. By Cartesian product, we produce a similar structure on any prism.
A prism complex X is a cellular complex obtained by identifying a collection of prisms along their faces by isometries (a face of a prism is a prism of smaller dimension). Given two points x,y\in X, a chain \Sigma between x and y is a sequence of points x_{1},\ldots,x_{n} such that x_{1}=x, x_{n}=y, and for every 1\leq i\leq n1, x_{i} and x_{i+1} belong to a same prism P_{i}. (Notice that a chain between two points of X always exists if X is connected: for instance, consider a path in the oneskeleton between two prisms containing our points.) The length of our chain is defined by \ell(\Sigma)=\sum\limits_{i=1}^{n1}d_{P_{i}}(x_{i},x_{i+1}). Finally, we introduce a pseudodistance on X by
d:(x,y)\mapsto\inf\{\ell(\Sigma)\mid\Sigma\ \text{chain between $x$ and $y$}\}. 
This is the standard way to define a metric on a polyhedral complex, as explained in [BH99, Definition 7.4]. From now on, we will suppose that every prism complex is endowed with this pseudometric. By defining a quasimedian complex as a prism complex obtained from a quasimedian graph by filling in every clique with a simplex and every 1skeleton of an ncube with an ncube, we want to prove:
Theorem 2.120.
Quasimedian complexes are CAT(0).
The following lemma will be fundamental in the proof of Theorem 2.120.
Lemma 2.121.
Let X be a quasimedian complex and x,y\in X two points. Let p_{x},p_{y} denote the minimal prisms containing x,y respectively. For every chain \Sigma between x and y, there exists another chain \Sigma^{*} between x and y such that \ell(\Sigma^{*})\leq\ell(\Sigma) which is included into the gated hull of \{p_{x},p_{y}\}.
Proof.
Let H denote the gated hull of \{p_{x},p_{y}\} and \Sigma=(x_{1},\ldots,x_{n}). A hyperplane J is bad relatively to \Sigma if J does not separate x and y, and \Sigma is not contained into the sector J^{+} delimited by J which contains x and y. Let b(\Sigma) denote the number of hyperplanes which are bad relatively to \Sigma. We want to prove that there exists a chain \Sigma^{*} between x and y satisfying b(\Sigma^{*})=0 and \ell(\Sigma^{*})\leq\ell(\Sigma). If b(\Sigma)=0, it suffices to set \Sigma^{*}=\Sigma. From now on, suppose that b(\Sigma)\geq 1, and let J be a bad hyperplane relatively to \Sigma which maximizes the distance to H. Let \partial=N(J)\cap J^{+}\subset\partial J.
Let (x_{r},\ldots,x_{s}) be a maximal subsegment of \Sigma included into N(J). We claim that the points x_{r},x_{s} belong to \partial.
Suppose by contradiction that x_{r}\notin\partial. Let P be a prism containing x_{r1} and x_{r}. Because x_{r1}\notin N(J), P\nsubseteq N(J); but P\cap N(J)\neq\emptyset because \Sigma is a chain. Let v\in P\cap N(J) be a vertex, J_{1},\ldots,J_{m} the hyperplanes dual to P, and, for every 1\leq i\leq m, v_{i}\in P a vertex adjacent to v and separated from it by J_{i}. If J,J_{1},\ldots,J_{m} are pairwise transverse, we deduce from Fact 2.75 that the gated hull of \{x,x_{1},\ldots,x_{m}\} defines a prism whose dual hyperplanes are J,J_{1},\ldots,J_{m}, hence P\subset N(J), a contradiction. Therefore, there exists a hyperplane J^{\prime} dual to P which is disjoint from J. On the other hand, because x_{r}\notin\partial and P\nsubseteq N(J), the prism P, and a fortiori the hyperplane J^{\prime}, must be included into a sector delimited by J which is different from J^{+}, the sector containing H. Consequently, we have d(J^{\prime},H)>d(J,H), contradicting our choice of J. Thus, we have proved that necessarily x_{r}\in\partial. Similarly, we show that x_{s}\in\partial.
From Lemma 2.28, we know that N(J) is naturally isomorphic to the product F(J)\times C, where C is a clique dual to J and F(J) the main fiber of J; moreover, there exists c\in C such that \partial=F(J)\times\{c\}. From \Sigma, we construct a new chain \Sigma^{\prime} by replacing the subsegment (x_{r},\ldots,x_{s}) with its projection onto \partial. It is worth noticing that any hyperplane intersecting \Sigma^{\prime} must intersect \Sigma, hence b(\Sigma^{\prime})\leq b(\Sigma); moreover, \ell(\Sigma^{\prime})\leq\ell(\Sigma). Finally, if we construct the chain \Sigma^{\prime\prime} from \Sigma by iterating this process with all the maximal subsegments of \Sigma included into N(J), then J will not be a bad hyperplane relatively to \Sigma^{\prime\prime}, hence b(\Sigma^{\prime\prime})\leq b(\Sigma)1. Moreover, \ell(\Sigma^{\prime\prime})\leq\ell(\Sigma).
By iterating this construction, we find a new chain \Sigma^{*} between x and y such that b(\Sigma^{*})=0 and \ell(\Sigma^{*})\leq\ell(\Sigma). The fact that b(\Sigma^{*})=0 precisely means that, for every hyperplane J which does not separate x and y, \Sigma^{*} must be included into the sector delimited by J which contains x and y. From Lemma 2.69, we deduce that \Sigma^{*}\subset H. ∎
The next step is to prove Theorem 2.120 in a special case.
Lemma 2.122.
Cubically finite quasimedian complexes are complete CAT(0) spaces.
Proof.
We argue by induction on the number of hyperplanes. A quasimedian graph without hyperplanes being a single vertex, there is nothing to prove in this case. If the hyperplanes of our quasimedian complex are pairwise transverse, we deduce from Lemma 2.74 that it is a prism, so that it must be a complete CAT(0) space as a Cartesian product of complete CAT(0) spaces. From now on, we suppose that our quasimedian complex X contains at least two disjoint hyperplanes J,J^{\prime}. Let J^{+} denote the sector delimited by J which contains our second hyperplane J^{\prime}.
Let Z denote the union of N(J) with all the sectors delimited by J which are different from J^{+}, and \partial the connected component of \partial J equal to N(J)\cap J^{+}=Z\cap J^{+}. Notice that
Claim 2.123.
For every vertices x\in Z and y\in J^{+},
d(x,y)=\inf\limits_{z\in\partial}\left(d(x,z)+d(z,y)\right). 
Let \Sigma=(x_{1},\ldots,x_{n}) be a chain from x to y. Let x_{k} denote the last element of the sequence x_{1},\ldots,x_{n} which belongs to Z. Notice that, since x_{n}=y\notin Z, necessarily k\leq n1, so that x_{k+1} is welldefined, and it belongs to J^{+}. Let P denote the minimal prism containing both x_{k} and x_{k+1}. Because x_{k+1} does not belong to Z, and that Z is a gated subgraph, there must exist a hyperplane J^{\prime} separating x_{k+1} from Z; a fortiori, J and J^{\prime} are disjoint. On the other hand, if x_{k} does not belong to \partial, J must separate x_{k} and x_{k+1} since they belong to distinct sectors delimited by J. Thus, the two disjoint hyperplanes J and J^{\prime} are transverse to the same prism P, which is impossible. Therefore, x_{k}\in\partial. We deduce that, if \Sigma_{1}=(x_{1},\ldots,x_{k}) and \Sigma_{2}=(x_{k},\ldots,x_{n}) are subchains of \Sigma, then
\ell(\Sigma)=\ell(\Sigma_{1})+\ell(\Sigma_{2})\geq d(x,x_{k})+d(x_{k},y)\geq% \inf\limits_{z\in\partial}\left(d(x,z)+d(z,y)\right). 
A fortiori, d(x,y)\geq\inf\limits_{z\in\partial}\left(d(x,z)+d(z,y)\right). The reverse inequality follows from the triangle inequality, which proves our claim.
Next, noticing that J is disjoint from J^{+} and that J^{\prime} is disjoint from Z, we deduce that the numbers of hyperplanes of Z and J^{+} are strictly less than the number of hyperplanes of X. Notice also that Z is a gated subcomplex of X according to Fact 2.103, so that it is a quasimedian complex on its own right. Therefore, we can apply our induction hypothesis to deduce that J^{+} and Z are CAT(0). In order to conclude that X is CAT(0), it is sufficient to show that \partial is a complete convex subspace in both Z and J^{+}, so that the Gluing Theorem [BH99, Theorem II.11.1] applies. (Indeed, Claim 2.123 states that the pseudodistance d coincides with the pseudodistance defined on the gluing Z\sqcup_{\partial}J^{+}.)
From Lemma 2.29, we already know that \partial is a gated subcomplex. In particular, it is a quasimedian complex on its own right, and we deduce that it is a complete CAT(0) space by applying our induction hypothesis since J being disjoint from \partial implies that the number of hyperplanes of \partial is strictly less than the number of hyperplanes of X. The convexity of \partial in Z and J^{+} follows from Lemma 2.121. Indeed, if we view \partial as a subcomplex of Z (resp. J^{+}), for every points x,y\in\partial, the unique geodesic in Z (resp. J^{+}) between x and y must be contained in the gated hull of \{p_{x},p_{y}\}, which in its turn must be included into \partial since \partial is gated. ∎
Corollary 2.124.
Quasimedian complexes are geodesic metric spaces.
Proof.
Let X be a quasimedian complex and x,y\in X two points. Let Y denote the gated hull of \{p_{x},p_{y}\}. According to Lemma 2.121,
\begin{array}[]{lcl}d_{X}(x,y)&=&\inf\{\ell(\Sigma)\mid\Sigma\ \text{chain % between $x$ and $y$}\}\\ &=&\inf\{\ell(\Sigma)\mid\Sigma\subset Y\ \text{chain between $x$ and $y$}\}\\ &=&d_{Y}(x,y).\end{array} 
On the other hand, since Y is a cubically finite quasimedian complex according to Corollary 2.71, it follows from Lemma 2.122 that Y is a CAT(0) space. Thus, the unique geodesic between x and y in Y produces a geodesic in X. ∎
Proof of Theorem 2.120..
Let X be a quasimedian complex and x,y,z\in X three points defining a geodesic triangle T. Let Y denote the gated hull of \{p_{x},p_{y},p_{z}\}. It follows from Lemma 2.121 that Y is convex, hence T\subset Y. On the other hand, according to Corollary 2.71, Y is a cubically finite quasimedian complex, so that T must satify the CAT(0) inequality since Y is CAT(0) according to Lemma 2.122. ∎
2.12 Locally quasimedian prism complexes
In [Gro87, Paragraph 4.2.C], Gromov noticed that a (finitedimensional) cube complex defines a CAT(0) space if and only if it is simply connected and if the links of its vertices are simplicial flag complexes. Because this condition on the links turns out to be often easy to verify, CAT(0) cube complexes have become a convenient source of CAT(0) spaces. In this section, we are interested in finding a similar criterion for determining whether a prism complex is or not quasimedian.
Definition 2.125.
Let X be a prism complex and v\in X a vertex. The simplicial part \mathrm{link}_{\triangle}(v) of the link \mathrm{link}(v) is the subcomplex generated by the edges associated to 2simplices of X; and the cubical part \mathrm{link}_{\square}(v) of the link \mathrm{link}(v) is the subcomplex generated by the edges associated to 2cubes of X.
Roughly speaking, our criterion can be thought of as follows: the cubical part of the prism complex must behave like a CAT(0) cube complex; its simplicial part must be a union of simplices such that the intersection between two such simplices is at most a single vertex; and the cubical and simplicial parts must interact in a “nonpositively curved” way, meaning that the interior of a prism is always filled in. Before stating the criterion we will be interested in, let us mention that, following [BCC{}^{+}13a], we define a flag simplicial complex as a simplicial complex in which every cycle of length three bounds a 2simplex, and a flag prism complex as a prism complex in which every cycle of length three bounds a 2simplex and every induced cycle of length four bounds a 2cube.
Definition 2.126.
A prism complex X is locally quasimedian if it is flag and if, for every vertex v\in X,

•
\mathrm{link}_{\square}(v) is a simplicial flag complexes;

•
\mathrm{link}_{\triangle}(v) is a disjoint union of simplices;

•
a simplex and a cube both containing v and intersecting along (at least) an edge must be contained into a prism.
Our main criterion is the following. The difficult part of the proof, namely showing that the 1skeleton of a simply connected locally quasimedian complex is a weakly modular graph, is contained in [BCC{}^{+}13a].
Theorem 2.127.
A prism complex is quasimedian if and only if it locally quasimedian and simply connected.
Proof.
Let X be a prism complex. Suppose that X is quasimedian. It is flag by definition and, because X is a CAT(0) space, according to Theorem 2.120, it must be simply connected (and even contractible). Next, let v\in X be a vertex. It follows from Fact 2.75 that \mathrm{link}_{\square}(v) is a flag complex, and because the intersection between two distinct cliques is either empty or a single vertex, according to Lemma 2.11, we also know that \mathrm{link}_{\triangle}(v) is a disjoint union of simplices. Finally, if C is a simplex and Q a cube such that both contain v and such that their intersection contains an edge, then C and Q must be included into the carrier of the hyperplane dual to the clique underlying C, so that it follows from Lemma 2.28 that C and Q generate a prism.
Conversely, suppose that X is locally quasimedian and simply connected. Then X is a bucolic complex, as defined in [BCC{}^{+}13a], and it follows from [BCC{}^{+}13a, Theorem 1] that the 1skeleton X^{(1)} is a weakly modular graph. Moreover, because cycles of length three in X bound 2simplices and because the simplicial parts of links are disjoint unions of simplices, K_{4}^{} cannot be an induced subgraph of X^{(1)}; and similarly, because induced cycles of length four in X bound 2cubes and because the cubical parts of links are simplicial complexes, K_{2,3} cannot be an induced subgraph of X^{(1)}. Therefore, X^{(1)} is a quasimedian graph, and a fortiori X is a quasimedian prism complex. ∎
3 Metrizing quasimedian graphs
In this section, we show that, if each clique C of a quasimedian graph X is endowed with a metric \delta_{C}, in such a way that the collection of all these metrics is coherent, then there exists a global metric \delta extending them. Moreover, if a group G acts on X and if our collection of metrics is Ginvariant, then the action G\curvearrowright X induces an isometric action G\curvearrowright(X,\delta). (The existence such collection of metrics will be studied in Section 5.) The main idea is that the global geometry of (X,\delta) reduces to the local geometries of the (C,\delta_{C})’s.
3.1 Canonical bijections
Before dealing with metrics on quasimedian graphs, we define a canonical bijection t_{C\to C^{\prime}}:C\to C^{\prime} between any two cliques C,C^{\prime} dual to the same hyperplane. This family of bijections will be fundamental not only in this section but in the whole article.
Definition 3.1.
Let X be a quasimedian graph and C,C^{\prime} two cliques dual to the same hyperplane. The canonical bijection from C to C^{\prime}, denoted by t_{C\to C^{\prime}}, is the restriction of the projection \mathrm{proj}_{C^{\prime}}:X\to C^{\prime} to C.
Alternatively, if J denotes the hyperplane dual to C and C^{\prime}, then Lemma 2.28 yields an isomorphism \Psi:N(J)\to F(J)\times C, and conjugating t_{C\to C^{\prime}} by \Psi produces the natural bijection (C,x)\mapsto(C^{\prime},x). Loosely speaking, t_{C\to C^{\prime}} translates C along J to C^{\prime}.
Now, we register some basic facts about the canonical bijections for future use.
Lemma 3.2.
For any two cliques C,C^{\prime} dual to the same hyperplane,
\mathrm{proj}_{C^{\prime}}=t_{C\to C^{\prime}}\circ\mathrm{proj}_{C}. 
The following observation will be needed to prove Lemma 3.2.
Fact 3.3.
Let X be a quasimedian graph, C a clique and x\in X a vertex. If D denotes the sector delimited by the hyperplane dual to C which contains x, then D\cap C=\{\mathrm{proj}_{C}(x)\}.
Proof.
Let J denote the hyperplane dual to C. According to Lemma 2.34, any hyperplane separating x and its projection onto C must be disjoint from C, so J cannot separate x and \mathrm{proj}_{C}(x). It follows that x and \mathrm{proj}_{C}(x) belong to the same sector delimited by J, hence \mathrm{proj}_{C}(x)\in D\cap C. On the other hand, any two vertices of C are separated by J, so D\cap C contains a single vertex. The conclusion follows. ∎
Proof of Lemma 3.2..
Let J denote the hyperplane dual to C,C^{\prime}. Fix a vertex x\in X and let D_{x} denote the sector delimited by J containing x. As a consequence of Fact 3.3, x and \mathrm{proj}_{C}(x) belong to D_{x}. Next, by applying Fact 3.3 twice, we know that
\{\mathrm{proj}_{C^{\prime}}(x)\}=D_{x}\cap C^{\prime}=\{\mathrm{proj}_{C^{% \prime}}\circ\mathrm{proj}_{C}(x)\}. 
Therefore,
\mathrm{proj}_{C^{\prime}}=\mathrm{proj}_{C^{\prime}}\circ\mathrm{proj}_{C}=t_% {C\to C^{\prime}}\circ\mathrm{proj}_{C}, 
which concludes the proof. ∎
Lemma 3.4.
If C,C^{\prime},C^{\prime\prime} are three cliques dual to the same hyperplane, then
t_{C^{\prime}\to C^{\prime\prime}}\circ t_{C\to C^{\prime}}=t_{C\to C^{\prime% \prime}}. 
Proof.
By applying Lemma 3.2 twice, we deduce that
t_{C\to C^{\prime\prime}}=\mathrm{proj}_{C^{\prime\prime}}\circ t_{C\to C^{% \prime}}=t_{C^{\prime}\to C^{\prime\prime}}\circ\mathrm{proj}_{C^{\prime}}% \circ t_{C\to C^{\prime}}. 
On the other hand, it is clear that \mathrm{proj}_{C^{\prime}}\circ t_{C\to C^{\prime}}=t_{C\to C^{\prime}}, so that the conclusion follows. ∎
Corollary 3.5.
If C,C^{\prime} are two cliques dual to the same hyperplane, then
t_{C\to C^{\prime}}^{1}=t_{C^{\prime}\to C}. 
Proof.
By applying the previous lemma, we know that
t_{C^{\prime}\to C}\circ t_{C\to C^{\prime}}=t_{C\to C}=\mathrm{id}_{C}. 
The conclusion follows. ∎
Remark 3.6.
As a consequence of Lemma 3.4, if C_{1},\ldots,C_{n} is a sequence of cliques all dual to the same hyperplanes, where C_{1}=C=C_{n} for some clique C, then
t_{C_{n1}\to C_{n}}\circ t_{C_{n2}\to C_{n1}}\circ\cdots\circ t_{C_{2}\to C% _{3}}\circ t_{C_{1}\to C_{2}}=\mathrm{id}_{C}. 
This monodromy condition satisfied by the canonical bijections generalizes the fact that, in a CAT(0) cube complex, hyperplanes are twosided.
3.2 Extending metrics
Given a quasimedian graph X, a system of (pseudo)metrics is the data, for every clique C of X, of a (pseudo)metric \delta_{C} defined on C. If C and C^{\prime} are two cliques dual to the same hyperplane, we define a new metric on C^{\prime} by
\delta_{C\to C^{\prime}}:(x,y)\longmapsto\delta_{C}\left(t_{C^{\prime}\to C}(x% ),t_{C^{\prime}\to C}(y)\right), 
where t_{C\to C^{\prime}}:C\to C^{\prime} is the canonical bijection defined in Section 3.1. Our system of (pseudo)metrics is coherent if \delta_{C^{\prime}}=\delta_{C\to C^{\prime}} for every pair of cliques C,C^{\prime} dual to the same hyperplane. In this section, our goal is to extend such a collection of (pseudo)metrics to a global one. In fact, several natural extensions exist, depending on some parameter p\in[1,+\infty] we fix.
Given two vertices x,y\in X, a chain \Sigma between x and y is a sequence of vertices
x_{0}=x,\ x_{1},\ldots,\ x_{n1},\ x_{n}=y 
such that, for every 0\leq i\leq n1, x_{i} and x_{i+1} belong to a common prism P_{i}=C^{i}_{1}\times\cdots\times C^{i}_{n(i)}. The \ell^{p}length of \Sigma is defined by
\ell_{p}(\Sigma)=\sum\limits_{i=0}^{n1}\delta^{p}_{P_{i}}(x_{i},x_{i+1}), 
where \delta^{p}_{P_{i}} denotes the \ell^{p}distance of the product (C^{i}_{1},\delta_{C^{i}_{1}})\times\cdots\times(C^{i}_{n(i)},\delta_{C^{i}_{n% (i)}}), ie.
\delta_{P_{i}}^{p}:\left((y_{k}),(z_{k})\right)\mapsto\left(\sum\limits_{k=1}^% {n(i)}\delta_{C_{k}^{i}}(y_{k},z_{k})^{p}\right)^{1/p} 
if p<+\infty and
\delta_{P_{i}}^{\infty}:((y_{k}),(z_{k}))\mapsto\max\limits_{1\leq k\leq n(i)}% \delta_{C_{k}^{i}}(y_{k},z_{k}) 
otherwise. It is worth noticing that the quantity \delta_{P_{i}}^{p}(x_{i},x_{i+1}) does not depend on the choice of the prism P_{i} containing both x_{i} and x_{i+1}, and that the metric \delta_{P_{i}}^{p} does not depend on the way we decomposed P_{i} as a product of cliques, because our system of (pseudo)metrics is coherent. Finally, we define the pseudodistance
\delta^{p}(x,y)=\inf\{\ell_{p}(\Sigma)\mid\Sigma\ \text{chain between $x$ and % $y$}\}. 
Notice that, for every vertices x,y\in X, the quantity \delta^{p}(x,y) is necessarily finite. Indeed, fix a geodesic z_{1},\ldots,z_{n} from x to y. A fortiori, for every 1\leq i\leq n1, the vertices z_{i} and z_{i+1} are adjacent, so that there exists a unique clique C_{i} containing them. Therefore, z_{1},\ldots,z_{n} defines a chain \Sigma from x to y, hence
\delta^{p}(x,y)\leq\ell_{p}(\Sigma)=\left(\sum\limits_{i=1}^{n1}\delta_{C_{i}% }(z_{i},z_{i+1})^{p}\right)^{1/p}<+\infty 
if p is finite, and
\delta^{\infty}(x,y)\leq\ell_{\infty}(\Sigma)=\max\limits_{1\leq i\leq n1}% \delta_{C_{i}}(z_{i},z_{i+1})<+\infty 
otherwise. Thus, we have defined a global pseudodistance \delta^{p}. Now, we want to prove that \delta^{p} turns out to be a metric (if we started from a collection of distances) which extends the \delta_{C}’s.
Proposition 3.7.
Let X be a quasimedian graph endowed with a coherent system of metrics. The global pseudometric \delta^{p} is a distance. Moreover, for every gated subgraph Y, the inclusion Y\subset X induces an isometric embedding (Y,\delta^{p}_{Y})\hookrightarrow(X,\delta^{p}) where \delta_{Y}^{p} denotes the global metric on Y obtained by extending the (restriction to Y of) our system of metrics.
The second part of the statement is contained in the following lemma, which follows by reproducing word for word the proof of Lemma 2.121.
Lemma 3.8.
Let X be a quasimedian graph endowed with a coherent system of metrics. Let x,y\in X be two vertices. For every chain \Sigma between x and y, there exists another chain \Sigma^{*} between x and y satisfying \ell(\Sigma^{*})\leq\ell(\Sigma) which is contained into the gated hull of \{x,y\}.
Proof of Proposition 3.7..
Let x,y\in X be two distinct vertices. A fortiori, there must exist a hyperplane J separating them. Set
\delta_{J}(x,y)=\delta_{C}(\mathrm{proj}_{C}(x),\mathrm{proj}_{C}(y)), 
where C is a clique dual to J. Notice that \delta_{J}(x,y)>0 because J separates x and y. Moreover, this quantity does not depend on the choice of C. Indeed, if C and C^{\prime} are two cliques dual to the same hyperplane, then we deduce from Lemma 3.2 that
\delta_{C}(p(x),p(y))=\delta_{C}(t\circ q(x),t\circ q(y))=\delta_{C\to C^{% \prime}}(q(x),q(y))=\delta_{C^{\prime}}(q(x),q(y)), 
where t=t_{C^{\prime}\to C}:C^{\prime}\to C is the canonical bijection from C to C^{\prime}, p:X\to C the projection onto C and q:X\to C^{\prime} the projection onto C^{\prime}. Now, let \Sigma=(x_{1},\ldots,x_{n}) be a chain between x and y, and, for every 1\leq i\leq n1, let P_{i} denote the minimal prism containing both x_{i} and x_{i+1}. Because J separates x and y, the set of prisms \{P_{m_{1}},\ldots,P_{m_{r}}\} intersected by J is non empty (for convenience, suppose that m_{1}<\cdots<m_{r}). Notice that x and x_{m_{1}} are not separated by J, so that p(x)=p(x_{m_{1}}); similarly, p(y)=p(x_{m_{r}+1}). Therefore,
\begin{array}[]{lcl}\ell_{p}(\Sigma)&=&\sum\limits_{i=1}^{n1}\delta_{P_{i}}^{% p}(x_{i},x_{i+1})\geq\sum\limits_{k=1}^{r1}\delta_{P_{m_{k}}}^{p}(x_{m_{k}},x% _{m_{k}+1})\geq\sum\limits_{k=1}^{r1}\delta_{C}(p(x_{m_{k}}),p(x_{m_{k}+1}))% \\ \\ &\geq&\delta_{C}(p(x_{m_{1}}),p(x_{m_{r}+1}))=\delta_{C}(p(x),p(y))=\delta_{J}% (x,y)\end{array} 
Because the quantity \delta_{J}(x,y) does not depend on the choice of the chain \Sigma, it follows that \delta^{p}(x,y)\geq\delta_{J}(x,y)>0. Thus, we have proved that \delta^{p} is a distance. The second statement of our proposition is a direct consequence of Lemma 3.8. ∎
So far, we have given infinitely many ways to extend a coherent system of metrics in order to get a global distance on the vertices of a quasimedian graph. Our next statement shows that, if the cubical dimension of our quasimedian graph is finite, then these global metrics are all Lipschitzequivalent. When the cubical dimension is infinite, the global geometry may be more interesting with respect to one of the possible metrics. For instance, in the case of CAT(0) cube complexes, we noticed in [Gen16b] that the CAT(0) cube complex associated to a C^{\prime}(1/4)T(4) polygonal complex is always hyperbolic with respect to the \ell^{\infty}distance, even when it is infinitedimensional (if so, it cannot be hyperbolic with respect to the \ell^{p}distance for any p\in[1,+\infty)).
Proposition 3.9.
Let X be a quasimedian graph endowed with a coherent system of metrics. If q\in[1,+\infty] satisfies \frac{1}{p}+\frac{1}{q}=1, then
\delta^{p}\leq\delta^{1}\leq\dim_{\square}(X)^{1/q}\cdot\delta^{p}. 
Proof.
Let x,y\in X be two vertices. Fix a chain \Sigma=(x_{1},\ldots,x_{n}) between x and y, and, for every 1\leq i\leq n1, let P_{i} denote a prism containing both x_{i} and x_{i+1}. Because p\geq 1, we know that \delta_{P_{i}}^{p}\leq\delta_{P_{i}}^{1} for every 1\leq i\leq n1, hence
\delta^{p}(x,y)\leq\ell_{p}(\Sigma)=\sum\limits_{i=1}^{n1}\delta_{P_{i}}^{p}(% x_{i},x_{i+1})\leq\sum\limits_{i=1}^{n1}\delta_{P_{i}}^{1}(x_{i},x_{i+1})=% \ell_{1}(\Sigma). 
A fortiori, \delta^{p}(x,y)\leq\delta^{1}(x,y). On the other hand, it follows from Hölder’s inequality that
\delta_{P_{i}}^{1}(x_{i},x_{i+1})\leq\dim_{\square}(P_{i})^{1/q}\cdot\delta_{P% _{i}}^{p}(x_{i},x_{i+1})\leq\dim_{\square}(X)^{1/q}\cdot\delta_{P_{i}}^{p}(x_{% i},x_{i+1}) 
for every 1\leq i\leq n1, hence
\delta^{1}(x,y)\leq\ell_{1}(\Sigma)\leq\dim_{\square}(X)^{1/q}\cdot\ell_{p}(% \Sigma). 
A fortiori, \delta^{1}(x,y)\leq\dim_{\square}(X)^{1/q}\cdot\delta^{p}(x,y). This concludes the proof. ∎
A natural problem is to determine how the (global) geometry of (X,\delta^{p}) reduces to the (local) geometries of the cliques (C,\delta_{C}). So we ask the following question:
Question 3.10.
Let X be a quasimedian graph endowed with a coherent system of metrics, and \mathcal{P} a property of metric spaces. If, for every clique C, the metric space (C,\delta_{C}) satisfies the property \mathcal{P}, when does (X,\delta^{p}) satisfies \mathcal{P} as well?
In this section and the next ones, we will answer this question for several properties, including being geodesic (Corollary 3.19), being locally finite (Lemma 3.25; see also Remark 3.26), and being CAT(0) (Proposition 3.11); Proposition 3.30 also studies the \ell^{p}compression of the global metric.
A good picture to keep in mind is that (X,\delta^{p}) is a kind of a “cubical agregate”. A CAT(0) cube complex is a union of cubes, which are products of edges, glued together in a “nonpositively curved” way. Similarly, (X,\delta^{p}) is a union of prisms, which are products of the spaces (C,\delta_{C}), glued together in a “nonpositively curved” way. From this analogy, it would not be surprising that any property characterizing a “nonpositively curved” behaviour transfers from local to global. We will motivate this idea by proving the following statement:
Proposition 3.11.
Let X be a quasimedian graph endowed with a coherent system of complete CAT(0) metrics. Then (X,\delta^{2}) is a CAT(0) space in which gated subgraphs of X are convex.
Proof.
First, let us prove by induction on the number of hyperplanes that, for any cubically finite gated subgraph Y\subset X, the metric space (Y,\delta^{2}) is a complete CAT(0) space. If Y has no hyperplane, it is a single vertex and there is nothing to prove; if Y has a single hyperplane, it is a single clique and there is nothing to prove. From now on, suppose that Y contains at least two hyperplanes. If the hyperplanes of Y are pairwise transverse, it follows from Lemma 2.74 that Y is a prism, so that (Y,\delta^{2}) is isometric to the \ell^{2}product of finitely many complete CAT(0) spaces. It follows from [BH99, Examples II.1.5] that (Y,\delta^{2}) is a complete CAT(0) space. Next, suppose that Y contains two disjoint hyperplanes, say J_{1} and J_{2}. Let Y_{2} denote the sector delimited by J_{1} containing J_{2} and Y_{1} the union of all the other sectors with N(J_{1}). According to Fact 2.103 and Proposition 2.15 respectively, Y_{1} and Y_{2} are gated. By applying our induction hypothesis, we know that (Y_{1},\delta^{2}) and (Y_{2},\delta^{2}) are complete CAT(0) spaces. Moreover, the intersection Y_{1}\cap Y_{2} is a fiber of J_{1}, which is also a gated subgraph according to Proposition 2.15, so we deduce from our induction hypothesis that Y_{1}\cap Y_{2} is a complete subspace of both (Y_{1},\delta^{2}) and (Y_{2},\delta^{2}); also, note that, as a consequence of Lemma 3.8, Y_{1}\cap Y_{2} is geodesic, and a fortiori convex since CAT(0) spaces are uniquely geodesic, in both (Y_{1},\delta^{2}) and (Y_{2},\delta^{2}). Next, notice that, by reproducing word for word the proof of Claim 2.123, we get the following general statement:
Fact 3.12.
Let X be a quasimedian graph endowed with a coherent system of metrics, and Y a gated subgraph containing two disjoint hyperplanes J_{1} and J_{2}. Let Y_{2} denote the sector delimited by J_{1} containing J_{2}, Y_{1} the union of all the other sectors with N(J_{1}), and \partial the fiber N(J_{1})\cap Y_{2} of J_{1}. Fix some p\in[1,+\infty]. Then
\delta^{p}(x,y)=\inf\limits_{z\in\partial}\left(\delta^{p}(x,z)+\delta^{p}(z,y% )\right) 
for every vertices x\in Y_{1} and y\in Y_{2}.
In particular, the equality
\delta^{2}(x,y)=\inf\limits_{z\in Y_{1}\cap Y_{2}}\left(\delta^{2}(x,z)+\delta% ^{2}(z,y)\right) 
holds for every vertices x\in Y_{1} and y\in Y_{2}. Therefore, it follows from the Gluing Theorem [BH99, Theorem II.11.1] that (Y,\delta^{2}) is a complete CAT(0) space.
Now, we can notice that (X,\delta^{2}) is a geodesic metric space. Indeed, the interval between two vertices is a cubically finite gated subgraph, and it follows from our previous observation that such an interval must be geodesic. On the other hand, such a geodesic defines a geodesic in (X,\delta^{2}) as a consequence of Lemma 3.8. Now, it makes sense to state that gated subgraphs of X are convex in (X,\delta^{2}), which is also a consequence of Lemma 3.8.
Finally, let \Delta=(x,y,z) be a geodesic triangle in (X,\delta^{2}). Since gated subgraphs are convex, necessarily \Delta must be included into the gated hull H of \{x,y,z\}. On the other hand, this hull is a cubically finite gated subgraph, so that the CAT(0) inequality must be satisfied in H, and a fortiori in (X,\delta^{2}). This concludes the proof. ∎
3.3 More on the global metric \delta^{1}
According to Proposition 3.9, our global metrics are all Lipschitz equivalent when the cubical dimension of the underlying quasimedian graph is finite, so that it is often possible to choose the metric which is the easiest to handle in the context we are interested in. In this section, our goal is to show that the global metric \delta^{1} can be described in a more explicit way, justifying the fact it will be our favorite metric most of the time. In all the article, we will set \delta=\delta^{1} for short.
Fix a quasimedian graph endowed with a coherent system of metrics. For every hyperplane J of X, we associate a pseudometric \delta_{J} on X defined by
\delta_{J}:(x,y)\mapsto\delta_{C}(\mathrm{proj}_{C}(x),\mathrm{proj}_{C}(y)), 
where C is any clique dual to J. Notice that \delta_{J} does not depend on the choice of the clique C since, whenever C and C^{\prime} are two cliques dual to the same hyperplane, we deduce from Lemma 3.2 that
\delta_{C}(p(x),p(y))=\delta_{C}(t\circ q(x),t\circ q(y))=\delta_{C\to C^{% \prime}}(q(x),q(y))=\delta_{C^{\prime}}(q(x),q(y)), 
where t=t_{C^{\prime}\to C}:C^{\prime}\to C is the canonical bijection from C to C^{\prime}, p:X\to C the projection onto C and q:X\to C^{\prime} the projection onto C^{\prime}. Notice that, for every vertices x,y\in X, the quantity \delta_{J}(x,y) is non zero if and only if J separates x and y.
Proposition 3.13.
Let X be a quasimedian graph endowed with a coherent system of metrics and x,y\in X two vertices. Then
\delta(x,y)=\sum\limits_{\text{$J$ hyperplane}}\delta_{J}(x,y)=\sum\limits_{% \text{$J$ separates $x$ and $y$}}\delta_{J}(x,y). 
Proof.
We want to prove by induction on k that, for any gated subgraph Y containing at most k hyperplanes, the equality
\delta(x,y)=\sum\limits_{\text{$J$ separates $x$ and $y$}}\delta_{J}(x,y) 
holds for every x,y\in Y.
Let Y be a gated subgraph with finitely many hyperplanes. If Y contains no hyperplane then Y is a single vertex and there is nothing to prove. And if the hyperplanes of Y are pairwise transverse, then Lemma 2.74 implies that Y is a single prism, so that the equality follows. From now on, suppose that Y contains at least two disjoint hyperplanes, say J_{1} and J_{2}. Let Y_{2} denote the sector delimited by J_{1} which contains J_{2}, and Y_{1} the union of all the other sectors with N(J_{1}). According to Fact 2.103 and Proposition 2.15 respectively, Y_{1} and Y_{2} are gated subgraphs. Moreover, the numbers of hyperplanes of Y_{1} and Y_{2} are smaller than the number of hyperplanes of Y, so that our induction hypothesis applies, ie.,
\delta(x,y)=\sum\limits_{\text{$J$ separates $x$ and $y$}}\delta_{J}(x,y) 
for every vertices x and y which both belong to either Y_{1} or Y_{2}. In particular, we know that the equality we want to prove holds if we consider two vertices of Y_{1} or two vertices of Y_{2}, so, in order to conclude, it is sufficient to verify our equality for two vertices x\in Y_{1} and y\in Y_{2}. On the other hand, if we denote by \partial the fiber N(J_{1})\cap Y_{2} of J_{1}, we know from Fact 3.12 that
\delta(x,y)=\inf\limits_{z\in\partial}\left(\delta(x,z)+\delta(z,y)\right). 
Fix some z\in\partial. Let x^{\prime},y^{\prime} denote respectively the projections of x,y onto \partial, and let (x^{\prime\prime},y^{\prime\prime},z^{\prime\prime}) be the quasimedian of the triple (x^{\prime},y^{\prime},z) (as defined in Section 2.7). Notice that, because \partial is a quasimedian graph on its own right, and because there exists a unique quasimedian in a quasimedian graph according to Proposition 2.84, necessarily x^{\prime\prime},y^{\prime\prime},z^{\prime\prime} belong to \partial.
For convenience, let us introduce the following notation. If a,b\in X are two vertices, we denote by \mathfrak{J}(a,b) the set of the hyperplanes separating a and b. Notice that, if c\in I(a,b), ie., if c lies on a geodesic between a and b, then \mathfrak{J}(a,b)=\mathfrak{J}(a,c)\sqcup\mathfrak{J}(c,b).
Because x^{\prime} is the projection of x onto \partial and that z\in\partial, necessarily x^{\prime}\in I(x,z). Moreover, by definition of a median triangle, there exists a geodesic between x^{\prime} and z passing through x^{\prime\prime} and z^{\prime\prime}. Therefore,
\mathfrak{J}(x,z)=\mathfrak{J}(x,x^{\prime})\sqcup\mathfrak{J}(x^{\prime},x^{% \prime\prime})\sqcup\mathfrak{J}(x^{\prime\prime},z^{\prime\prime})\sqcup% \mathfrak{J}(z^{\prime\prime},z). 
Similarly,
\mathfrak{J}(z,y)=\mathfrak{J}(z,z^{\prime\prime})\sqcup\mathfrak{J}(z^{\prime% \prime},y^{\prime\prime})\sqcup\mathfrak{J}(y^{\prime\prime},y^{\prime})\sqcup% \mathfrak{J}(y^{\prime},y) 
As a consequence,
\begin{array}[]{lcl}\delta(x,z)+\delta(z,y)&=&\displaystyle\sum\limits_{J\in% \mathfrak{J}(x,z)}\delta_{J}(x,z)+\sum\limits_{J\in\mathfrak{J}(z,y)}\delta_{J% }(z,y)\\ \\ &\geq&\displaystyle\sum\limits_{J\in\mathfrak{J}(x,x^{\prime})\sqcup\mathfrak{% J}(x^{\prime},x^{\prime\prime})}\delta_{J}(x,z)+\sum\limits_{J\in\mathfrak{J}(% x^{\prime\prime},z^{\prime\prime})}\delta_{J}(x,z)\\ &&\displaystyle+\sum\limits_{J\in\mathfrak{J}(z^{\prime\prime},y^{\prime\prime% })}\delta_{J}(z,y)+\sum\limits_{J\in\mathfrak{J}(y^{\prime\prime},y^{\prime})% \sqcup\mathfrak{J}(y^{\prime},y)}\delta_{J}(z,y)\end{array} 
Notice that, if J\in\mathfrak{J}(x,x^{\prime})\sqcup\mathfrak{J}(x^{\prime},x^{\prime\prime}), then z and y^{\prime\prime} are not separated by J. Indeed, if J\in\mathfrak{J}(x,x^{\prime}), then it follows from Lemma 2.34 that J separates x from \partial, but z and y^{\prime\prime} belongs to \partial, so J cannot separate them; and if J\in\mathfrak{J}(x^{\prime},x^{\prime\prime}), our claim follows from Fact 2.88. So
\sum\limits_{J\in\mathfrak{J}(x,x^{\prime})\sqcup\mathfrak{J}(x^{\prime},x^{% \prime\prime})}\delta_{J}(x,z)=\sum\limits_{J\in\mathfrak{J}(x,x^{\prime})% \sqcup\mathfrak{J}(x^{\prime},x^{\prime\prime})}\delta_{J}(x,y^{\prime\prime}), 
and similarly,
\sum\limits_{J\in\mathfrak{J}(y^{\prime\prime},y^{\prime})\sqcup\mathfrak{J}(y% ^{\prime},y)}\delta_{J}(z,y)=\sum\limits_{J\in\mathfrak{J}(y^{\prime\prime},y^% {\prime})\sqcup\mathfrak{J}(y^{\prime},y)}\delta_{J}(y^{\prime\prime},y). 
Next, notice that it follows from Fact 2.90 that \mathfrak{J}(x^{\prime\prime},z^{\prime\prime})=\mathfrak{J}(z^{\prime\prime},% y^{\prime\prime})=\mathfrak{J}(x^{\prime\prime},y^{\prime\prime}), so that
\begin{array}[]{lcl}\displaystyle\sum\limits_{J\in\mathfrak{J}(x^{\prime\prime% },z^{\prime\prime})}\delta_{J}(x,z)+\sum\limits_{J\in\mathfrak{J}(z^{\prime% \prime},y^{\prime\prime})}\delta_{J}(z,y)&=&\displaystyle\sum\limits_{J\in% \mathfrak{J}(x^{\prime\prime},y^{\prime\prime})}\left(\delta_{J}(x,z)+\delta_{% J}(z,y)\right)\\ \\ &\geq&\displaystyle\sum\limits_{J\in\mathfrak{J}(x^{\prime\prime},y^{\prime% \prime})}\delta_{J}(x,y)=\sum\limits_{J\in\mathfrak{J}(x^{\prime\prime},y^{% \prime\prime})}\delta_{J}(x,y^{\prime\prime})\end{array} 
The last equality is justified by the fact that no hyperplane of \mathfrak{J}(x^{\prime\prime},y^{\prime\prime}) separates y and y^{\prime\prime}. Indeed, a hyperplane J separating y and y^{\prime\prime} separates either y and y^{\prime} or y^{\prime} and y^{\prime\prime}. In the former case, it follows from Lemma 2.34 that J separates y from \partial, so that it cannot separate y and y^{\prime\prime} since y^{\prime\prime}\in\partial; and in the latter case, the conclusion is a consequence of Fact 2.88. By noticing that
\mathfrak{J}(x,x^{\prime})\sqcup\mathfrak{J}(x^{\prime},x^{\prime\prime})% \sqcup\mathfrak{J}(x^{\prime\prime},y^{\prime\prime})=\mathfrak{J}(x,y^{\prime% \prime})\ \text{and}\ \mathfrak{J}(y^{\prime\prime},y^{\prime})\sqcup\mathfrak% {J}(y^{\prime},y)=\mathfrak{J}(y^{\prime\prime},y), 
we finally get
\begin{array}[]{lcl}\delta(x,z)+\delta(z,y)&\geq&\displaystyle\sum\limits_{J% \in\mathfrak{J}(x,x^{\prime})\sqcup\mathfrak{J}(x^{\prime},x^{\prime\prime})% \sqcup\mathfrak{J}(x^{\prime\prime},y^{\prime\prime})}\delta_{J}(x,y^{\prime% \prime})+\sum\limits_{J\in\mathfrak{J}(y^{\prime\prime},y^{\prime})\sqcup% \mathfrak{J}(y^{\prime},y)}\delta_{J}(y^{\prime\prime},y)\\ \\ &\geq&\displaystyle\sum\limits_{J\in\mathfrak{J}(x,y^{\prime\prime})}\delta_{J% }(x,y^{\prime\prime})+\sum\limits_{J\in\mathfrak{J}(y^{\prime\prime},y)}\delta% _{J}(y^{\prime\prime},y)=\delta(x,y^{\prime\prime})+\delta(y^{\prime\prime},y)% \end{array} 
where the last equality is justified by the fact that x,y^{\prime\prime} both belong to Y_{1} and that y^{\prime\prime},y both belong to Y_{2}. Notice that y^{\prime\prime}\in I(x,y). Indeed, by definition of a quasimedian triangle, we know that there exists a geodesic [x^{\prime},y^{\prime}]\subset\partial between x^{\prime} and y^{\prime} passing through y^{\prime\prime}. On the other hand, a hyperplane separating x and x^{\prime} must separate x from \partial according to Lemma 2.34, and a fortiori from Y_{2}, so that it cannot intersect [x^{\prime},y^{\prime}] or separate y and y^{\prime}; and similarly, a hyperplane separating y and y^{\prime} cannot intersect [x^{\prime},y^{\prime}] or separate x and x^{\prime}. Therefore, fixing some geodesics [x,x^{\prime}] and [y,y^{\prime}] respectively between x and x^{\prime}, and y and y^{\prime}, we deduce that the concatenation [x,x^{\prime}]\cup[x^{\prime},y^{\prime}]\cup[y^{\prime},y] is a geodesic since it cannot intersect a hyperplane twice. So we have proved that y^{\prime\prime} belongs to some geodesic between x and y, which precisely means that y^{\prime\prime} belongs to the interval I(x,y). A fortiori, y^{\prime\prime}\in\partial\cap I(x,y). Thus, we have proved that
\delta(x,y)=\inf\limits_{z\in\partial\cap I(x,y)}\left(\delta(x,z)+\delta(z,y)% \right). 
On the other hand, for every z\in\partial\cap I(x,y),
\delta(x,z)+\delta(z,y)=\sum\limits_{J\in\mathfrak{J}(x,z)}\delta_{J}(x,z)+% \sum\limits_{J\in\mathfrak{J}(z,y)}\delta_{J}(y,z). 
Notice that, for every J\in\mathfrak{J}(x,z), the projections of y and z onto N(J) coincide since J does not separate y and z, hence \delta_{J}(x,z)=\delta_{J}(x,y). Similarly, \delta_{J}(z,y)=\delta_{J}(x,y) for every J\in\mathfrak{J}(z,y). Thus,
\delta(x,z)+\delta(z,y)=\sum\limits_{J\in\mathfrak{J}(x,z)\sqcup\mathfrak{J}(z% ,y)}\delta_{J}(x,y)=\sum\limits_{J\in\mathfrak{J}(x,y)}\delta_{J}(x,y). 
It follows that
\delta(x,y)=\sum\limits_{J\in\mathfrak{J}(x,y)}\delta_{J}(x,y), 
concluding the proof. ∎
Example 3.14.
If \delta_{C}:(x,y)\mapsto\left\{\begin{array}[]{cl}0&\text{if}\ x=y\\ 1&\text{otherwise}\end{array}\right. for every clique C of X, then \delta=d.
We know from Proposition 3.7 that gated subgraphs define isometrically embedded subspaces of (X,\delta^{p}), for every p\in[1,+\infty]. When p=1, we are also able to show that they are convex in the following meaning:
Definition 3.15.
Let (S,d) be a metric space. The interval of two points x,y\in S is the set
I(x,y)=\{z\in X\mid d(x,y)=d(x,z)+d(z,y)\}. 
A subset R\subset S is convex if I(x,y)\subset R for every x,y\in R.
It is worth noticing that, if (S,d) is a geodesic metric space, then I(x,y) is the union of all the geodesics between x and y; and a subset R\subset S is convex if and only if every geodesic between two points of R stays in R, which is the usual definition of convexity for geodesic metric spaces.
Proposition 3.16.
Let X be a quasimedian graph endowed with a coherent system of metrics. Every gated subgraph of X is convex in (X,\delta).
This proposition is essentially a consequence of the following lemma, which is an adaptation of Lemma 2.121.
Lemma 3.17.
Let X be a quasimedian graph endowed with a coherent system of metrics. Let x,y,z\in X be three vertices such that z does not belong to the gated hull of \{x,y\}. There exists some \epsilon(z)>0 such that, for every chain \Sigma between x and y passing through z, there exists another chain \Sigma^{*} between x and y satisfying \ell(\Sigma^{*})\leq\ell(\Sigma)\epsilon(z).
Proof.
Let J be a hyperplane separating z from the gated hull H of \{x,y\} which maximizes the distance to H. Notice that z belongs to N(J) since otherwise there would exist a hyperplane separating z from N(J), contradicting the maximality of J. Let J^{+} denote the sector delimited by J which contains H, and set \partial=N(J)\cap J^{+}\subset\partial J.
Let (x_{r},\ldots,x_{s}) be the maximal subsegment of \Sigma included into N(J) which contains z. We claim that the points x_{r},x_{s} belong to \partial.
Suppose by contradiction that x_{r}\notin\partial. Let P be a prism containing x_{r1} and x_{r}. Because x_{r1}\notin N(J), P\nsubseteq N(J); but P\cap N(J)\neq\emptyset because \Sigma is a chain. Let v\in P\cap N(J) be a vertex, J_{1},\ldots,J_{m} the hyperplanes dual to P, and, for every 1\leq i\leq m, v_{i}\in P a vertex adjacent to v and separated from it by J_{i}. If J,J_{1},\ldots,J_{m} are pairwise transverse, we deduce from Fact 2.75 that the gated hull of \{x,x_{1},\ldots,x_{m}\} defines a prism whose dual hyperplanes are J,J_{1},\ldots,J_{m}, hence P\subset N(J), a contradiction. Therefore, there exists a hyperplane J^{\prime} dual to P which is disjoint from J. On the other hand, because x_{r}\notin\partial and P\nsubseteq N(J), the prism P, and a fortiori the hyperplane J^{\prime}, must be included into a sector delimited by J which is different from J^{+}, the sector containing H. Consequently, we have d(J^{\prime},H)>d(J,H), contradicting our choice of J. Thus, we have proved that necessarily x_{r}\in\partial. Similarly, we show that x_{s}\in\partial.
Let \Sigma^{\prime} denote the subchain (x_{r},\ldots,x_{s}). For every r\leq i\leq s1, fix a prism P_{i}\subset N(J) containing both x_{i} and x_{i+1}. Without loss of generality, we can suppose that P_{i} contains a clique dual to J, so that P_{i} decomposes a product C_{i}\times P_{i}^{\prime} where P_{i}^{\prime} is a prism included into \partial and C_{i} a clique dual to J. Notice that, if m:X\to\partial denotes the projection onto \partial, then (m(x_{r}),\ldots,m(x_{s})) defines a chain since m(x_{i}) and m(x_{i+1}) both belong to P_{i}^{\prime} for every r\leq i\leq s1; let \Sigma^{\prime\prime} denote this new chain. Now, if we denote by q:X\to C the projection onto some clique C dual to J and q_{i}:X\to C_{i} the projection onto C_{i} for every r\leq i\leq s, we have
\begin{array}[]{lcl}\ell(\Sigma^{\prime})&=&\displaystyle\sum\limits_{i=r}^{s% 1}\delta_{P_{i}}(x_{i},x_{i+1})=\sum\limits_{i=r}^{s1}\left(\delta_{P_{i}^{% \prime}}(m(x_{i}),m(x_{i+1}))+\delta_{C_{i}}(q_{i}(x_{i}),q_{i}(x_{i+1}))% \right)\\ \\ &=&\displaystyle\ell(\Sigma^{\prime\prime})+\sum\limits_{i=r}^{s1}\delta_{C}(% q(x_{i}),q(x_{i+1}))\end{array} 
On the other hand, we know by assumption that there exists some r+1\leq k\leq s1 satisfying x_{k}=z, so that
\sum\limits_{i=r}^{s1}\delta_{C}(q(x_{i}),q(x_{i+1}))\geq\sum\limits_{i=r}^{k% 1}\delta_{C}(q(x_{i}),q(x_{i+1}))\geq\delta_{C}(q(x_{r}),q(x_{k}))=\delta_{C}% (q(x),q(z)). 
Therefore, \ell(\Sigma^{\prime})\geq\ell(\Sigma^{\prime\prime})+\epsilon(z) where
\epsilon(z)=\min\{\delta_{J}(x,z)\mid\text{$J$ separates $z$ from $\{x,y\}$}\}. 
Notice that \epsilon(z)>0 since there exists only finitely many hyperplanes separating z from \{x,y\}. A fortiori, if \Sigma^{*} denote the chain obtained from \Sigma by replacing \Sigma^{\prime} with \Sigma^{\prime\prime},
\ell(\Sigma^{*})=\ell(\Sigma)+\ell(\Sigma^{\prime\prime})\ell(\Sigma^{\prime}% )\leq\ell(\Sigma)\epsilon(z). 
This concludes the proof. ∎
Proof of Proposition 3.16..
Let Y be a gated subgraph and x,y\in Y two vertices. We want to prove that, if z\in X is a vertex which does not belong to Y, then z\notin I(x,y). This will prove that I(x,y) must be included into Y, concluding the proof of the proposition.
Let \epsilon(z) denote the constant given by Lemma 3.17. Fix a chain \Sigma_{1} (resp. \Sigma_{2}) between x and z (resp. between z and y) such that \ell(\Sigma_{1})\delta(x,z)\leq\epsilon(z)/4 (resp. \ell(\Sigma_{2})\delta(z,y)\leq\epsilon(z)/4). A fortiori, the concatenation of \Sigma_{1} and \Sigma_{2} defines a chain \Sigma between x and y passing through z. According to Lemma 3.17, there exists a chain \Sigma^{*} between x and y such that \ell(\Sigma^{*})\leq\ell(\Sigma)\epsilon(z). We have
\begin{array}[]{lcl}\delta(x,y)&\leq&\ell(\Sigma^{*})\leq\ell(\Sigma)\epsilon% (z)\\ \\ &\leq&\ell(\Sigma_{1})+\ell(\Sigma_{2})\epsilon(z)\leq\delta(x,z)+\frac{% \epsilon(z)}{4}+\delta(z,y)+\frac{\epsilon(z)}{4}\epsilon(z)\\ \\ &\leq&\delta(x,z)+\delta(z,y)\frac{\epsilon(z)}{2}<\delta(x,z)+\delta(z,y)% \end{array} 
Therefore, z does not belong to the interval I(x,y). ∎
Now, fix a quasimedian graph X endowed with a coherent system of geodesic metrics. A natural problem is to determine whether (X,\delta) is geodesic as well, and if so, to determine (at least some of) its geodesics. For this purpose, let us define a broken geodesic \gamma=\gamma_{1}\cup\cdots\cup\gamma_{n} between two vertices x,y\in X as the data of

•
a sequence of cliques C_{1},\cdots,C_{n} such that x\in C_{1}, y\in C_{n} and such that C_{i}\cap C_{i+1} reduces to a single vertex x_{i} for every 1\leq i\leq n1;

•
a sequence of paths \gamma_{1}\subset C_{1},\ldots,\gamma_{n}\subset C_{n} such that \gamma_{i} is a geodesic in (C_{i},\delta_{C_{i}}) between x_{i} and x_{i+1} for every 1\leq i\leq n1, with the convention that x_{0}=x and x_{n}=y;
such that x_{0},x_{1},\ldots,x_{n1},x_{n} defines a geodesic in (X,d) between x and y.
It is worth noticing that any two vertices of X are linked by a broken geodesic. Indeed, fix two vertices x,y\in X and consider some geodesic x_{0}=x,\ x_{1},\ldots,x_{n1},\ x_{n}=y between x and y in (X,d). For every 0\leq i\leq n1, let C_{i} denote the unique clique of X containing the edge (x_{i},x_{i+1}) and \gamma_{i} a geodesic between x_{i} and x_{i+1} in (C_{i},\delta_{C_{i}}). Then
\gamma=\gamma_{0}\cup\gamma_{1}\cdots\cdots\cup\gamma_{n1}\cup\gamma_{n} 
defines a broken geodesic between x and y.
Lemma 3.18.
A broken geodesic defines a geodesic in (X,\delta).
Proof.
Let \gamma=\gamma_{0}\cup\gamma_{1}\cup\cdots\gamma_{n} be a broken geodesic between two vertices x,y\in X. For every 0\leq i\leq n1, let C_{i} denote the clique containing \gamma_{i} and x_{i},x_{i+1}\in C_{i} the endpoints of \gamma_{i}. By definition,
\varphi:x_{0}=x,\ x_{1},\ldots,x_{n1},\ x_{n}=y 
is a geodesic between x and y in (X,d). We claim that \gamma defines a geodesic in (X,\delta).
Let a,b\in\gamma. If there exists some 0\leq i\leq n such that a,b\in\gamma_{i}, then it is clear that \delta(a,b)=\delta_{C_{i}}(a,b) is equal to the length of the subsegment of \gamma between a and b. So let us suppose that there exist i<j such that a\in\gamma_{i}\backslash\{x_{i+1}\} and b\in\gamma_{j}\backslash\{x_{j}\}. Set
\overline{\varphi}:a,\ x_{i+1},\ x_{i+2},\ldots,x_{j1},\ x_{j},\ b. 
Notice that \overline{\varphi} defines a geodesic in (X,d). Indeed, the clique containing (a,x_{i+1}) is C_{i}, the clique containing (x_{j},b) is C_{j}, and finally, for every i+1\leq r\leq j1, the clique containing (x_{r},x_{r+1}) is C_{r}, so that no hyperplane intersects \overline{\varphi} twice (otherwise, a hyperplane would intersect \varphi twice, which is impossible). As a consequence, the hyperplanes separating a and b are precisely the hyperplanes dual to the cliques mentionned above, hence
\delta(a,b)=\delta_{C_{i}}(a,x_{i+1})+\sum\limits_{r=i+1}^{j1}\delta_{C_{r}}(% x_{r},x_{r+1})+\delta_{C_{j}}(x_{j},b). 
By noticing that this is precisely the length of the subsegment of \gamma between a and b, we conclude that \gamma is a geodesic. ∎
Thus, combined with Proposition 3.16, we deduce from this lemma that:
Corollary 3.19.
Let X be quasimedian graph endowed with a coherent system of geodesic metrics. Then (X,\delta) is a geodesic metric space in which the gated subgraphs are convex.
Remark 3.20.
In the statements of Lemma 3.18 and Corollary 3.19, we consider geodesic metric spaces which are either all discrete (ie., with geodesics defined on \mathbb{N}) or all continuous (ie., with geodesics defined on \mathbb{R}). More precisely, Lemma 3.18 proves that a broken geodesic constructed from discrete (resp. continuous) geodesics defines a discrete (resp. continuous) geodesic, and Corollary 3.19 states that the global metric associated to a system of discrete (resp. continuous) geodesic spaces is discrete (resp. continuous) geodesic.
In order to motivate the idea that studying (X,\delta^{1}) may be useful to deduce properties of (X,\delta^{p}) when the cubical dimension of X is finite, let us notice that, as a consequence of Proposition 3.9, (X,\delta^{1}) is proper if and only if so is (X,\delta^{p}). This observation allows us to prove the following criterion, which is however purely technical and will be used only for the proof of Theorem 7.7.
Lemma 3.21.
Let Y be a quasimedian graph of finite cubical dimension endowed with a coherent system of complete and proper CAT(0) metrics. Suppose that Y contains a convex subgraph X satisfying the following conditions:

•
every vertex of X is contained into finitely many cliques of X;

•
every vertex of Y belongs to a clique containing an edge of X;

•
there exists some constant K>0 such that, for any clique C and for any distinct vertices x,y\in X, we have \delta_{C}(x,y)\geq K.
Then (Y,\delta^{p}) is proper.
Proof.
According to Proposition 3.9, it is sufficient to prove that (Y,\delta^{1}) is proper. We first want to prove that any ball of (Y,\delta^{1}) contains only finitely many points of X. Notice that, because any vertex of Y is adjacent to some vertex of X, we may suppose without loss of generality that our ball is centered at some point of X.
Let B(x,R) be a ball of radius R (with respect to \delta^{1}) centered at some point x\in X. Suppose that y\in B(x,R)\cap X. By considering some broken geodesic between x and y, provided by Lemma 3.18, we find a geodesic x_{1},\ldots,x_{r} between x and y in Y (with respect to its combinatorial metric) such that, if C_{i} denotes the clique containing x_{i} and x_{i+1} for every 1\leq i\leq r1, then \delta_{C_{i}}(x_{i},x_{i+1})\geq K (notice that x_{i}\in X for every 1\leq i\leq r by convexity of X). In particular,
R\geq\delta^{1}(x,y)=\sum\limits_{i=1}^{r1}\delta_{C_{i}}(x_{i},x_{i+1})\geq Kr, 
hence r\leq R/K. Thinking of y as a variable point, because there exist only finitely many cliques of X containing x, one has only finitely many choices on C_{1}, and because (C_{1},\delta_{C_{1}}) is proper and that \delta_{C_{1}}(x,x_{1})\geq K, one has only finitely many choices on x_{1} as a point of C_{1}. Therefore, there are finitely many possible choices on x_{1}. Similarly, x_{1} being fixed, one has only finitely many choices on C_{2} since there exist only finitely many cliques of X containing x_{1}, and, because (C_{2},\delta_{C_{2}}) is proper and that \delta_{C_{2}}(x_{1},x_{2}), on has only finitely many choices on x_{2} as a point of X_{2}. Therefore, there are finitely many possible choices on x_{2}. By iterating the argument at most R/K times, we conclude that there exist only finitely many possible y’s, ie., B(x,R)\cap X is finite.
Now, we are ready to conclude the proof of our lemma. So, fixing some bounded sequence (y_{n}) of vertices (with respect to \delta^{1}), we want to find a converging subsequence (with respect to \delta^{1}). For every n\geq 0, let C_{n} be a clique of X such that the unique clique of Y containing C_{n} contains y_{n}. If B denotes a ball containing all the y_{n}’s, because we proved that B\cap X is finite necessarily (y_{n}) eventually lies inside some C_{m} up to taking a subsequence. Since (C_{m},\delta_{C_{m}}) is proper, we conclude that this subsequence must contain a converging subsequence, which concludes the proof. ∎
For future use, we record the first claim of the previous proof:
Claim 3.22.
Under the hypotheses of Lemma 3.21, any ball of (Y,\delta^{p}) contains finitely many vertices of X.
3.4 Actions of groups
In this section, we are interested in group actions. First notice that, if a group G acts on a quasimedian graph X, endowed with a system of (pseudo)metrics which is coherent and Ginvariant (ie., \delta_{gC}(gx,gy)=\delta_{C}(x,y) for every clique C and every vertices x,y\in C), then G acts on (X,\delta). Indeed, if we fix two vertices x,y\in X and an element g\in G, then, by denoting J_{1},\ldots,J_{n} the hyperplanes separating x and y and by fixing some clique C_{i} dual to J_{i} for every 1\leq i\leq n, then
\begin{array}[]{lcl}\delta(gx,gy)&=&\sum\limits_{i=1}^{n}\delta_{gC_{i}}(% \mathrm{proj}_{gC_{i}}(gx),\mathrm{proj}_{gC_{i}}(gy))=\sum\limits_{i=1}^{n}% \delta_{gC_{i}}(g\cdot\mathrm{proj}_{C_{i}}(x),g\cdot\mathrm{proj}_{C_{i}}(y))% \\ \\ &=&\sum\limits_{i=1}^{n}\delta_{C_{i}}(\mathrm{proj}_{C_{i}}(x),\mathrm{proj}_% {C_{i}}(y))=\delta(x,y)\end{array} 
Now, we would like to determine when the induced action G\curvearrowright(X,\delta) is metrically proper or geometric. We give some criteria below.
Lemma 3.23.
Let G be a group acting on a quasimedian graph X endowed with a Ginvariant coherent system of metrics. Suppose that

•
X contains finitely many Gorbits of cliques;

•
for every clique C, the action \mathrm{stab}(C)\curvearrowright(C,\delta_{C}) is cobounded.
Then G acts coboundedly on (X,\delta).
Proof.
Let C_{1},\ldots,C_{n} be a collection of cliques so that any clique of X is a translate of some C_{i}. For every 1\leq i\leq n, fix a vertex x_{i}\in C_{i} and let A_{i} be the diameter of a fundamental domain containing x_{i} for the action \mathrm{stab}(C_{i})\curvearrowright(C_{i},\delta_{C_{i}}). Set A=\max\limits_{1\leq i\leq n}A_{i} and B=\max\limits_{2\leq i\leq n}\delta(x_{1},x_{i}). For every x\in X, there exists some g\in G and 1\leq i\leq n such that gx\in C_{i}. Then, there exists some h\in\mathrm{stab}(C_{i}) such that \delta(x_{i},hgx)\leq A. Therefore,
\delta(x_{1},x)\leq\delta(x_{1},x_{i})+\delta(x_{i},hgx)\leq A+B. 
As a consequence, the ball of radius A+B centered at x_{1} defines a fundamental domain for the action G\curvearrowright(X,\delta). A fortiori, this action is cobounded. ∎
Lemma 3.24.
Let G be a group acting on a quasimedian graph X endowed with a Ginvariant coherent system of metrics. Suppose that

•
any vertex of X belongs to finitely many cliques;

•
vertexstabilisers are finite;

•
for every clique C, the space (C,\delta_{C}) is locally finite;

•
suppose that the metrics \delta_{C} are uniformly discrete, ie., there exists some constant K>0 such that, for any clique C and any points x,y\in C, \delta_{C}(x,y)\geq K.
Then G acts metrically properly on (X,\delta).
Essentially, this result will be a consequence of the following lemma:
Lemma 3.25.
Let X be a quasimedian graph endowed with a coherent system of metrics. Suppose that

•
any vertex of X belongs to finitely many cliques;

•
for every clique C, the space (C,\delta_{C}) is locally finite;

•
suppose that the metrics \delta_{C} are uniformly discrete, ie., there exists some constant K>0 such that, for any clique C and any points x\neq y\in C, \delta_{C}(x,y)\geq K.
Then (X,\delta) is locally finite.
Proof.
Up to rescaling all the \delta_{C}’s, we may suppose without loss of generality that K=1. As a consequence, d\leq K\cdot\delta=\delta.
We want to prove that, for every x\in X and every R\geq 0, the ball B_{\delta}(x,R) is finite; noticing that
B_{\delta}(x,R+1)\subset\bigcup\limits_{y\in B_{\delta}(x,R)}B_{\delta}(y,1), 
it is sufficient to prove that B_{\delta}(x,1) is finite for every x\in X.
Notice that, for every y\in B_{\delta}(x,1), we have d(x,y)\leq\delta(x,y)\leq 1, ie., x and y belong to a common clique. Let C_{1},\ldots,C_{n} denote the cliques containing x. Notice that, for every 1\leq i\leq n and every y\in C_{i}, we have \delta(x,y)=\delta_{C_{i}}(x,y). Consequently,
B_{\delta}(x,1)\subset\bigcup\limits_{i=1}^{n}B_{\delta_{C_{i}}}(x,1), 
so that we deduce that B_{\delta}(x,1) is finite since the balls B_{\delta_{C_{i}}}(x,1) are themselves finite by local finiteness of the spaces (C_{i},\delta_{C_{i}}). ∎
Remark 3.26.
In the statement of Lemma 3.25, the second condition is of course necessary. However, the first and third conditions are not necessary. Nevertheless, the three conditions turn out to define a necessary and sufficient criterion for the local finiteness of (X,\delta) if a group acts on X with finitely many orbits of cliques.
Proof of Lemma 3.24..
Let x\in X and R\geq 0. For every y\in B_{\delta}(x,R), fix some g_{y}\in G such that g_{y}\cdot x=y; if no such element of G exists, set g_{y}=1. Then, set F=\{g_{y}\mid y\in B_{\delta}(x,R)\}. Notice that
\#F\leq\#\mathcal{B}_{\delta}(x,R)<+\infty 
according to the previous lemma. Now, if g\in G satisfies d_{\delta}(x,gx)\leq R, then there exists some f\in F such that gx=fx, hence g\in F\cdot\mathrm{stab}(x). Thus, we have proved that
\{g\in G\mid d_{\delta}(x,gx)\leq R\}\subset F\cdot\mathrm{stab}(x). 
On the other hand, we know that F is finite and \mathrm{stab}(x) is finite by hypothesis. Therefore, G acts metrically properly on (X,\delta). ∎
Corollary 3.27.
Let G be a group acting on a quasimedian graph X endowed with a Ginvariant coherent system of discrete metrics. Suppose that:

•
any vertex of X belongs to finitely many cliques;

•
vertexstabilisers are finite;

•
X contains finitely many Gorbits of cliques;

•
for every clique C, the action \mathrm{stab}(C)\curvearrowright(C,\delta_{C}) is geometric.
Then G acts geometrically on (X,\delta).
3.5 \ell^{p}compression
Roughly speaking, the \ell^{p}compression of a metric space X is a real number in [0,1] which quantifies how much it is necessary to deform the geometry of X in order to embed it in some L^{p}space; a precise definition is given below. In this section, our goal is to find a lower bound on the \ell^{p}compression of the global metric associated to some coherent system of metrics on a quasimedian graph with respect to the \ell^{p}compressions of the local metrics.
Definition 3.28.
Let f:X\to Y be a Lipschitz map between two metric spaces. The compression of f, denoted by \alpha(f), is the supremum of the \alpha’s such that there exists some constant C>0 so that the inequality
C\cdot d(x,y)^{\alpha}\leq d(f(x),f(y)) 
holds for every x,y\in X. Given a metric space X and some p\geq 1, the \ell^{p}compression of X is the supremum of the \alpha(f)’s where f is a Lipschitz map from X to a L^{p}space. In particular, if G is a finitely generated group, its \ell^{p}compression is defined as the \ell^{p}compression of G endowed with the word metric associated to a finite generating set; notice that the choice of this generating set does not modify the compression since any two word metrics (with respect to two finite generating sets) are Lipschitzequivalent.
It is worth noticing that, if X is a uniformly discrete metric space (ie., there exists some B>0 such that d(x,y)\geq B for every distinct points x,y\in X), it is possible to define alternatively the \ell^{p}compression of X as the supremum of the \alpha’s such that there exist a Lipschitz map f:X\to L to some L^{p}space such that
C\cdot d(x,y)^{\alpha}D\leq d(f(x),f(y)) 
for some C,D>0 and for all x,y\in X; see for instance [Dre11, Lemma 2.1]. As a consequence, the \ell^{p}compression of a finitely generated group turns out to be a quasiisometric invariant.
Definition 3.29.
Let X be a quasimedian graph endowed with a coherent system of metrics. A system of Lipschitz \ell^{p}maps is the data, for each clique C of X, of one Lipschitz map f_{C}:(C,\delta_{C})\to L_{C} from the clique C to some L^{p}space L_{C}. Such a system is coherent if

•
L_{C}=L_{C^{\prime}} if C and C’ are two cliques dual to the same hyperplane;

•
for every two cliques C,C^{\prime} of X dual to the same hyperplane and for every two vertices x,y\in C,
f_{C}(x,y)=f_{C^{\prime}}(t_{C\to C^{\prime}}(x),t_{C\to C^{\prime}}(y)) where t_{C\to C^{\prime}}:C\to C^{\prime} is the canonical bijection.
The main result of this section is the following.
Proposition 3.30.
Let X be a quasimedian graph endowed with a coherent system of uniformly discrete metrics \{(C,\delta_{C})\mid C\ \text{clique}\}. Suppose that the collection \{(C,\delta_{C})\mid\text{$C$ clique}\} contains finitely many isometry classes. Then
\alpha_{p}(X,\delta)\geq\min\left(\frac{1}{p},\inf\limits_{C~{}\text{clique}}% \alpha_{p}(C,\delta_{C})\right). 
Proof.
First of all, notice that rescaling all the \delta_{C}’s by a common positive constant does not modify the compressions \alpha_{p}(X,\delta) and \alpha_{p}(C,\delta_{C}). Therefore, since our collection of metrics is uniformly discrete, we can suppose without loss of generality that \delta_{C}(x,y)\geq 1 for every clique C and every distinct vertices x,y\in C. Next, if \min\limits_{\text{$C$ clique}}\alpha_{p}(C,\delta_{C})=0, there is nothing to prove, so we suppose that \min\limits_{\text{$C$ clique}}\alpha_{p}(C,\delta_{C})>0.
Fix some collection \mathcal{Q} of cliques of X such that each hyperplane contains exactly one clique of \mathcal{Q}, and some 0<\epsilon<\min\limits_{\text{$C$ clique}}\alpha_{p}(C,\delta_{C}). For every clique C\in\mathcal{Q}, fix a Lipschitz map f_{C} from (C,\delta_{C}) to some L^{p}space L_{C} such that \alpha(f_{C})\geq\alpha_{p}(C,\delta_{C})\epsilon. Because \{(C,\delta_{C})\mid C\in\mathcal{Q}\} contains only finitely many isometry classes, we can choose the maps f_{C} so that, for every positive \eta<\min\limits_{C\in\mathcal{Q}}\alpha_{p}(C,\delta_{C}), there exist constants A_{\eta},B>0 (which do not depend on C) such that
A_{\eta}\cdot\delta_{C}(x,y)^{\alpha(f_{C})\eta}\leq\f_{C}(x)f_{C}(y)\\leq B% \cdot\delta_{C}(x,y) 
for every C\in\mathcal{Q} and every x,y\in C. Now, if C is an arbitrary clique of X, the hyperplane dual to C must contain a unique clique Q of \mathcal{Q}. We set L_{C}=L_{Q} and f_{C}=f_{Q}\circ t_{C\to Q}, in order to define a system of Lipschitz \ell^{p}maps, which is coherent by construction; notice that, because our system of metrics is coherent, the canonical bijection t_{C\to Q}:(C,\delta_{C})\to(Q,\delta_{Q}) defines an isometry, so that, for every positive \eta<\min\limits_{C\in\mathcal{Q}}\alpha_{p}(C,\delta_{C}), the inequalities
A_{\eta}\cdot\delta_{C}(x,y)^{\alpha(f_{Q})\eta}\leq\f_{C}(x)f_{C}(y)\\leq B% \cdot\delta_{C}(x,y) 
hold for every clique C labelled by Q\in\mathcal{Q} and every vertices x,y\in C. In particular,
\alpha(f_{C})=\alpha(f_{Q})\geq\alpha_{p}(Q,\delta_{Q})\epsilon=\alpha_{p}(C,% \delta_{C})\epsilon. 
For convenience, let p_{Q}:X\to Q denote the projection onto Q for every Q\in\mathcal{Q}. Define
f:\left\{\begin{array}[]{ccc}X&\to&L=\bigoplus\limits_{C\in\mathcal{Q}}^{\ell^% {p}}L_{C}\\ x&\mapsto&(f_{C}(p_{C}(x))f_{C}(p_{C}(x_{0})))_{C}\end{array}\right., 
where x_{0}\in X is a basepoint we fix.
Claim 3.31.
f is welldefined, Lipschitz and satisfies
\alpha(f)\geq\min\left(\frac{1}{p},\min\limits_{C~{}\text{clique}}\alpha(f_{C}% )\right). 
First, let us verify that f is welldefined. For every vertex x\in X, only finitely many hyperplanes separate x and x_{0}, so that p_{C}(x) and p_{C}(x_{0}) differ only for finitely many C\in\mathcal{Q}; a fortiori, f(x) belongs to L. Next, in order to justify that f is a Lipschitz map, notice that, for every vertices x,y\in X, we have
\begin{array}[]{lcl}\f(x)f(y)\&=&\displaystyle\left(\sum\limits_{C\in% \mathcal{Q}}\f_{C}(p_{C}(x))f_{C}(p_{C}(y))\^{p}\right)^{1/p}\\ \\ &\leq&\displaystyle B\cdot\left(\sum\limits_{C\in\mathcal{Q}}\delta_{C}(p_{C}(% x),p_{C}(y))^{p}\right)^{1/p}\\ \\ &\leq&\displaystyle B\cdot\sum\limits_{C\in\mathcal{Q}}\delta_{C}(p_{C}(x),p_{% C}(y))=B\cdot\delta(x,y)\end{array} 
The final step is to understand the compression of f. For convenience, we set c=\min\left(\frac{1}{p},\min\limits_{C~{}\text{clique}}\alpha(f_{C})\eta\right) for some fixed positive \eta<\min\limits_{\text{$C$ clique}}\alpha_{C}. For every vertices x,y\in X,
\begin{array}[]{lcl}\f(x)f(y)\^{1/c}&=&\displaystyle\left(\sum\limits_{C\in% \mathcal{Q}}\f_{C}(p_{C}(x))f_{C}(p_{C}(y))\^{p}\right)^{1/pc}\\ \\ &\geq&\displaystyle A_{\eta}^{1/c}\cdot\left(\sum\limits_{C\in\mathcal{Q}}% \delta_{C}(p_{C}(x),p_{C}(y))^{p(\alpha(f_{C})\eta)}\right)^{1/pc}\\ \\ &\geq&\displaystyle A_{\eta}^{1/c}\cdot\left(\sum\limits_{C\in\mathcal{Q}}% \delta_{C}(p_{C}(x),p_{C}(y))^{pc}\right)^{1/pc}\\ \\ &\geq&\displaystyle A_{\eta}^{1/c}\cdot\sum\limits_{Q\in\mathcal{Q}}\delta_{C}% (p_{C}(x),p_{C}(y))=A_{\eta}^{1/c}\cdot\delta(x,y)\end{array} 
Thus, we have shown that \alpha(f)\geq\min\left(\frac{1}{p},\min\limits_{C~{}\text{clique}}\alpha(f_{C}% )\eta\right), and because the inequality holds for every sufficently small \eta>0, this concludes the proof of our claim. Because
\alpha_{p}(X,\delta)\geq\alpha(f)\geq\min\left(\frac{1}{p},\min\limits_{C~{}% \text{clique}}\alpha(f_{C})\right)\geq\min\left(\frac{1}{p},\min\limits_{C~{}% \text{clique}}\alpha(C,\delta_{C})\epsilon\right) 
is true for every \epsilon>0, our proposition follows. ∎
4 Cubulating quasimedian graphs
In this section, we show that, if each clique C of a quasimedian graph X is endowed with a collection of walls \mathcal{W}(C), in such a way that the family of all these wallspaces is coherent, then there exists a global collection of walls \mathcal{HW} extending them. Moreover, if a group G acts on X and if our collection of wallspaces is Ginvariant, then the action G\curvearrowright X induces an action G\curvearrowright(X,\mathcal{HW}). (The existence such collection of wallspaces will be studied in Section 5.) The main point is that the global combinatorics of (X,\mathcal{HW}) reduces to the local combinatorics of the (C,\mathcal{W}(C))’s. In the final subsection, we mention how this formalism can be adapted to some generalisations of spaces with walls, namely measured wallspaces and spaces with labelled partitions.
The rest of this introduction is used to fixed the definitions and notation related to spaces with walls.
Given a set X, a wall is a partition W=\{h,h^{c}\} where h\notin\{\emptyset,X\}; we will refer to h and its complementary h^{c} as the halfspaces of W. If x,y\in X are two points, we say that W separates x and y if they belong to distinct halfspaces of W. Finally, given a collection of walls \mathcal{W}, we say that (X,\mathcal{W}) is a space with walls if any two points of X are separated by only finitely many walls. Notice that this condition allows us to define a pseudodistance d_{\mathcal{W}} on X by counting the number of walls separating two given points.
In particular, spaces with walls are examples of spaces with partitions as defined in Section 2.4 (see Remark 2.66). In our definition of spaces with walls, we allow duplicates, ie., two walls may be indistinguishable in the sense that they induce the same partition on X; compare with [HW12, Remark 2.2]. We cubulate spaces with walls by quasicubulating them as spaces with partitions, as explained in Section 2.4. The main difference between our cubulation and the cubulation for instance described in [HW12] is that we define when two walls are nested and then we say that two walls which are not nested are transverse, whereas usually one says that two walls \{h_{1},h_{1}^{c}\} and \{h_{2},h_{2}^{c}\} are transverse if the four intersections
h_{1}\cap h_{2},\ h_{1}\cap h_{2}^{c},\ h_{1}^{c}\cap h_{2},\ h_{1}^{c}\cap h_% {2}^{c} 
are non empty and then one says that two walls which are not transverse are nested. Observe that if \mathcal{P}_{1} and \mathcal{P}_{2} are two partitions such that A_{1}\cap A_{2}\neq\emptyset for every A_{1}\in\mathcal{P}_{1} and A_{2}\in\mathcal{P}_{2}, then \mathcal{P}_{1} and \mathcal{P}_{2} are transverse, so that the usual definition of transversality implies ours. However, the converse does not hold in general, as illustrated by Figure 6.
Of course, given a space with walls (X,\mathcal{W}) and the CAT(0) cube complex C(X,\mathcal{W}) obtained by cubulating it, we are interested in reading the geometry of the CAT(0) cube complex C(X,\mathcal{W}) directly from (X,\mathcal{W}). For this purpose, we will essentially use Theorem 2.56 in order to link the combinatorics of the hyperplanes of C(X,\mathcal{W}) with the combinatorics of the walls of (X,\mathcal{W}). For instance, we define the dimension of (X,\mathcal{W}) as the maximal cardinality of a collection of pairwise transverse walls, and we show:
Lemma 4.1.
Let (X,\mathcal{W}) be a space with walls and let C(X,\mathcal{W}) denote the CAT(0) cube complex obtained by cubulating (X,\mathcal{W}). Then \dim(X,\mathcal{W})=\dim C(X,\mathcal{W}).
Proof.
The dimension of C(X,\mathcal{W}) is equal to the maximal cardinality of a collection of pairwise transverse hyperplanes of C(X,\mathcal{W}) (see for instance Proposition 2.73). Therefore, the bijection between the walls of \mathcal{W} and the hyperplanes of C(X,\mathcal{W}) given by Theorem 2.56 provides the conclusion. ∎
In the sequel, given a group G acting on a space with walls (X,\mathcal{W}), we will say that the action G\curvearrowright(X,\mathcal{W}) satisfies some property \mathcal{P} if the associated action of G on the CAT(0) cube complex C(X,\mathcal{W}) obtained by cubulating (X,\mathcal{W}) satisfies \mathcal{P} as well. In particular,
Lemma 4.2.
A group G acts metrically properly on a space with walls (X,\mathcal{W}) if and only if it acts metrically properly on the pseudometric space (X,d_{\mathcal{W}}).
Proof.
Let C(X,\mathcal{W}) denote the CAT(0) cube complex obtained by cubulating (X,\mathcal{W}). Fix some vertex x\in X and let \sigma_{x} denote the principal orientation associated. In particular, \sigma_{x} is a vertex of C(X,\mathcal{W}). Notice that, using Lemma 2.62, we know that
d_{C(X,\mathcal{W})}(\sigma_{x},g\cdot\sigma_{x})=d_{C(X,\mathcal{W})}(\sigma_% {x},\sigma_{g\cdot x})=d_{\mathcal{W}}(x,g\cdot x) 
for every g\in G. As a consequence, the action G\curvearrowright C(X,\mathcal{W}) is metrically proper if and only if the action G\curvearrowright(X,d_{\mathcal{W}}) is metrically proper as well. ∎
Lemma 4.3.
Let (X,\mathcal{W}) be a finite dimensional space with walls. A group G acts cocompactly on (X,\mathcal{W}) if and only if X contains finitely many Gorbits of maximal collections of pairwise transverse walls.
Proof.
Let C(X,\mathcal{W}) denote the CAT(0) cube complex obtained by cubulating (X,\mathcal{W}). Notice that, according to Lemma 4.1, C(X,\mathcal{W}) is finite dimensional. As a consequence, Proposition 2.79 applies. Therefore, G acts cocompactly on C(X,\mathcal{W}) if and only if C(X,\mathcal{W}) has finitely many Gorbits of maximal cubes if and only if C(X,\mathcal{W}) contains finitely many Gorbits of maximal collections of pairwise transverse hyperplanes. According to Theorem 2.56, the latter condition is equivalent to the following statement: X contains finitely many Gorbits of maximal collections of pairwise transverse walls. ∎
Such a result can also be formulated for special actions, as defined by Haglund and Wise in [HW08]. Let us adapt the definition for popsets (and, in particular, for spaces with partitions, including quasimedian graphs).
Definition 4.4.
Let G be a group acting on a popset (X,<,\mathfrak{P}). The action is special if none of the following pathological configurations happen:

•
there exist a partition \mathcal{P}\in\mathfrak{P} and an element g\in G such that \mathcal{P} and g\mathcal{P} are transverse;

•
there exist a partition \mathcal{P}\in\mathfrak{P} and an element g\in G such that \mathcal{P} and g\mathcal{P} are tangent;

•
there exist three partitions \mathcal{P}_{1},\mathcal{P}_{2},\mathcal{P}_{3}\in\mathfrak{P} and an element g\in G such that \mathcal{P}_{1} is transverse to \mathcal{P}_{2}, \mathcal{P}_{3} is tangent to \mathcal{P}_{2}, and g\mathcal{P}_{1}=\mathcal{P}_{3}.
Remark 4.5.
We emphasize that a group acting (geometrically and) specially on some CAT(0) cube complex may not be a special group as defined in [HW08]. For instance, it is proved in [HW99a] that there exists a nonpositivelycurved square of finite groups whose fundamental group is not residually finite. Such a group acts geometrically and specially on some CAT(0) square complex but is not (virtually) special. Nevertheless, a group acting geometrically and virtually specially on a CAT(0) cube complex turns out to be virtually special if it is residually finite, or more generally if it is virtually torsionfree.
Remark 4.6.
Usually, when defining special actions on CAT(0) cube complexes, the following additional condition is also required: if J is a hyperplane delimiting two halfspaces J^{+} and J^{}, then gJ^{+}\neq J^{} for every g\in\mathrm{stab}(J). This condition does not appear in the definition of specialness given in [HW08, Definition 3.2]. Nevertheless, it is worth noticing that, when our group turns out to be residually finite or virtually torsionfree (ie., when our group turns out to be special, which is the case which interests us), up to taking a finiteindex subgroup this condition is always satisfied [HW08, Proposition 3.10].
It clearly follows from Theorem 2.56 that the action on a popset is special if and only if the action on the associated quasimedian graph is special as well. In particular, acting specially on a space with walls implies acting specially on the dual CAT(0) cube complex.
Definition 4.7.
Let G be a group acting on a space with walls (X,\mathcal{W}). The obstruction to the specialness \mathrm{Obs}(G\curvearrowright(X,\mathcal{W})) is the set of the elements of G which appear in a pathological configuration.
We conclude this introduction by the following easy observation, which will be useful later.
Fact 4.8.
Let G be a group acting on a space with walls (X,\mathcal{W}). Suppose that G contains a subgroup H such that the induced action H\curvearrowright(X,\mathcal{W}) is special. If F is a set of representatives of G/H different from H, then
\mathrm{Obs}(G\curvearrowright(X,\mathcal{W}))\subset FH. 
Proof.
Since G=FH\sqcup H and \mathrm{Obs}(G\curvearrowright(X,\mathcal{W}))\cap H=\emptyset, because the action H\curvearrowright(X,\mathcal{W}) is special, the conclusion follows. ∎
4.1 Extending walls
Let X be a quasimedian graph. If W=\{h,h^{c}\} is a wall defined on some clique C, we extend it as a wall on X by
\overline{W}=\left\{\bigcup\limits_{x\in h}[C,x],\bigcup\limits_{x\in h^{c}}[C% ,x]\right\}=\left\{\mathrm{proj}_{C}^{1}(h),\mathrm{proj}_{C}^{1}(h^{c})% \right\}. 
Recall that, if x\in C, [C,x] denotes the sector delimited by the hyperplane dual to C which contains x. More generally, if \mathcal{W}(C) is a collection of walls of C, we introduce
\overline{\mathcal{W}}(C)=\left\{\overline{W}\mid W\in\mathcal{W}(C)\right\}. 
If C^{\prime} is a second clique such that C and C^{\prime} are dual to the same hyperplane, we are able to transfer a collection of walls of C to C^{\prime} by defining
\mathcal{W}(C\to C^{\prime})=\left\{t(W)=\{t(h),t(h^{c})\}\mid W=\{h,h^{c}\}% \in\mathcal{W}(C)\right\}, 
where t=t_{C\to C^{\prime}} is the canonical bijection from C to C^{\prime} as defined in Section 3.1. It is worth noticing that this operation does not modify the extended collection of walls:
Fact 4.9.
For any two cliques dual to the same hyperplane, the equality
\overline{\mathcal{W}}(C\to C^{\prime})=\overline{\mathcal{W}}(C) 
holds, where \overline{\mathcal{W}}(C\to C^{\prime})=\{\overline{W}\mid W\in\mathcal{W}(C% \to C^{\prime})\}.
Proof.
Let M\in\overline{\mathcal{W}}(C) be a wall. So there exists some W=\{h,h^{c}\}\in\mathcal{W}(C) such that M=\overline{W}. We deduce from Lemma 3.2 that
M=\left\{\mathrm{proj}_{C}^{1}(h),\mathrm{proj}_{C}^{1}(h^{c})\right\}=\left% \{\mathrm{proj}_{C^{\prime}}^{1}(t(h)),\mathrm{proj}_{C^{\prime}}^{1}(t(h^{c% }))\right\}\in\overline{\mathcal{W}}(C\to C^{\prime}), 
since \{t(h),t(h^{c})\}\in\mathcal{W}(C\to C^{\prime}). Conversely, consider a wall M\in\overline{\mathcal{W}}(C\to C^{\prime}). So there exists some W=\{t(h),t(h^{c})\}\in\mathcal{W}(C\to C^{\prime}), where \{h,h^{c}\}\in\mathcal{W}(C), such that M=\overline{W}. Once again, we deduce from Lemma 3.2 that
M=\left\{\mathrm{proj}_{C^{\prime}}^{1}(t(h)),\mathrm{proj}_{C^{\prime}}^{1}% (t(h^{c}))\right\}=\left\{\mathrm{proj}_{C}(h),\mathrm{proj}_{C}(h^{c})\right% \}\in\overline{\mathcal{W}}(C), 
since \{h,h^{c}\}\in\mathcal{W}(C). This concludes the proof. ∎
Definition 4.10.
Let X be a quasimedian graph. A system of wallspaces is the data of a set of walls \mathcal{W}(C) for each clique C, such that (C,\mathcal{W}(C)) defines a space with walls. It is coherent if \mathcal{W}(C^{\prime})=\mathcal{W}(C\to C^{\prime}) for every cliques C,C^{\prime} dual to the same hyperplane.
From now on, we fix a quasimedian graph X endowed with a coherent system of wallspaces. Notice that, if C and C^{\prime} are two cliques dual to the same hyperplane, then, because our system of wallspaces is coherent and thanks to Fact 4.9, we have
\overline{\mathcal{W}}(C^{\prime})=\overline{\mathcal{W}}(C\to C^{\prime})=% \overline{\mathcal{W}}(C). 
As a consequence, it makes sense to set, for every hyperplane J of X, \mathcal{W}(J)=\overline{\mathcal{W}}(C), where C is any clique dual to J. Thus, the coherent system of wallspaces of X defines a collection of walls, which we will refer to as hyperplanewalls, denoted by
\mathcal{HW}=\bigcup\limits_{\text{$J$ hyperplane of $X$}}\overline{\mathcal{W% }}(J). 
Notice that every clique is endowed with the pseudometric associated to its collection of walls. Our first result essentially states that the pseudodistance associated to \mathcal{HW} coincides with the global pseudometric obtained from this system of pseudometrics as explained in Section 3.3.
Lemma 4.11.
Let x,y\in X be two vertices. Let J_{1},\ldots,J_{n} denote the hyperplanes separating x and y, and, for every 1\leq i\leq n, fix a clique C_{i} dual to J_{i}. If p_{i}:X\to C_{i} denotes the projection onto C_{i} for every 1\leq i\leq n, then
d_{\mathcal{HW}}(x,y)=\sum\limits_{i=1}^{n}d_{\mathcal{W}(C_{i})}(p_{i}(x),p_{% i}(y)). 
Proof.
For every 1\leq i\leq n, let q_{i}:X\to N(J_{i}) denote the projection onto N(J_{i}). Because a hyperplanewall is a union of sectors delimited by some fixed hyperplane, the underlying hyperplane of a hyperplanewall separating x and y must separate x and y as well, hence
d_{\mathcal{HW}}(x,y)=\sum\limits_{J\ \text{hyperplane}}d_{\mathcal{W}(J)}(x,y% )=\sum\limits_{i=1}^{n}d_{\mathcal{W}(J_{i})}(x,y). 
Moreover, for every 1\leq i\leq n, the quantity d_{\mathcal{W}(J_{i})}(x,y) depends only on the sectors containing x and y, hence d_{\mathcal{W}(J_{i})}(x,y)=d_{\mathcal{W}(J_{i})}(q_{i}(x),q_{i}(y)). Noticing that p_{i}(q_{i}(x))=p_{i}(x) and similarly p_{i}(q_{i}(y))=p_{i}(y), according to Corollary 2.40, it is sufficient to prove the following fact in order to deduce that d_{\mathcal{W}(J_{i})}(x,y)=d_{\mathcal{W}(C_{i})}(p_{i}(x),p_{i}(y)), which concludes the proof of our lemma.
Fact 4.12.
Let J be a hyperplane, x,y\in N(J) two vertices and C a clique dual to J. If p:X\to C denotes the projection onto C, then d_{\mathcal{W}(J)}(x,y)=d_{\mathcal{W}(C)}(p(x),p(y)).
Let M\in\mathcal{W}(J). Because \mathcal{W}(J)=\overline{\mathcal{W}}(C), there exists a wall W=\{h,h^{c}\}\in\mathcal{W}(C) such that M=\overline{W}=\left\{p^{1}(h),p^{1}(h^{c})\right\}. Then M separates x and y if and only if W separates p(x) and p(y). Therefore d_{\mathcal{W}(J)}(x,y)=d_{\mathcal{W}(C)}(p(x),p(y)). ∎
The second step is to understand when two hyperplanewalls are transverse or tangent. This is the aim of our next two lemmas.
Lemma 4.13.
Let M_{1}\in\mathcal{W}(J_{1}) and M_{2}\in\mathcal{W}(J_{2}) be two hyperplanewalls. Then M_{1} and M_{2} are transverse if and only if either J_{1} and J_{2} are transverse, or J_{1}=J_{2} and there exist two transverse walls W_{1},W_{2}\in\mathcal{W}(C), where C is a clique dual to J_{1}=J_{2}, such that M_{1}=\overline{W_{1}} and M_{2}=\overline{W_{2}}.
Proof.
If J_{1} and J_{2} are nested, then, because the halfspaces of M_{1} and M_{2} are unions of sectors delimited by J_{1} and J_{2} respectively, some halfspace delimited by M_{1} or M_{2} must be strictly included into another halfspace delimited by the other wall. A fortiori, M_{1} and M_{2} are nested.
Suppose that J_{1} and J_{2} are transverse. According to Proposition 2.73, there exists a prism P, which is a product of two cliques C_{1},C_{2} dual to J_{1},J_{2} respectively. Let u be the single vertex of the intersection C_{1}\cap C_{2}, and, for i=1,2, let u_{i}\in C_{i} be a vertex such that M_{i} separates u and u_{i}. Let w be the vertex (u_{1},u_{2}) of P=C_{1}\times C_{2}. Moreover, M_{1} separates u_{2} and w, because u_{1} and w belong to the same sector delimited by J_{1}; similarly, M_{2} separates u_{1} and w. Thus, if M_{i}^{+} denotes the halfspace delimited by M_{i} which contains u, and M_{i}^{} its complement, then
u\in M_{1}^{+}\cap M_{2}^{+},\ u_{1}\in M_{1}^{}\cap M_{2}^{+},\ u_{2}\in M_{% 1}^{+}\cap M_{2}^{},\ w\in M_{1}^{}\cap M_{2}^{}. 
Therefore, M_{1} and M_{2} are transverse.
Finally, suppose that J_{1}=J_{2}. So, given a clique C dual to J_{1}=J_{2}, there exist two walls W_{1},W_{2}\in\mathcal{W}(C) such that M_{1}=\overline{W}_{1} and M_{2}=\overline{W}_{2}. Say W_{1}=\{h_{1},h_{1}^{c}\} and W_{2}=\{h_{2},h_{2}^{c}\}, so that, if p:X\to C denotes the projection onto C, we have M_{1}=\{p^{1}(h_{1}),p^{1}(h_{1}^{c})\} and M_{2}=\{p^{1}(h_{2}),p^{1}(h_{2}^{c})\}. Notice that the inclusion h_{1}\subset h_{2} is equivalent to p^{1}(h_{1})\subset p^{1}(h_{2}), and similarly the inclusion h_{2}\subset h_{1} is equivalent to p^{1}(h_{2})\subset p^{1}(h_{1}). As a consequence, M_{1} and M_{2} are nested if and only if W_{1} and W_{2} are nested as well; a fortiori, M_{1} and M_{2} are transverse if and only if W_{1} and W_{2} are transverse as well. ∎
Corollary 4.14.
Let X be a quasimedian graph endowed with a coherent system of wallspaces. Then \dim(X,\mathcal{HW})\leq\dim_{\square}(X)\cdot\sup\limits_{C\ \mathrm{clique}}% \dim(C,\mathcal{W}(C)).
Proof.
Let M_{1},\ldots,M_{n} be a collection of pairwise transverse hyperplanewalls. Let J_{1},\ldots,J_{m} denote the collection of pairwise distinct hyperplanes of X underlying the walls M_{1},\ldots,M_{n}. We deduce from Lemma 4.13 that J_{1},\ldots,J_{m} are pairwise transverse, hence m\leq\dim_{\square}(X). Moreover, if J_{j} is the underlying hyperplane of M_{i_{1}},\ldots,M_{i_{k}}, then, according to Lemma 4.13, there exist pairwise transverse walls W_{i_{1}},\ldots,W_{i_{k}}\in\mathcal{W}(C_{j}), where C_{j} is a clique dual to J_{j}, such that M_{i_{r}}=\overline{W_{i_{r}}} for every 1\leq r\leq k. Therefore, k\leq\dim(C_{j},\mathcal{W}(C_{j})). The conclusion follows. ∎
Lemma 4.15.
Suppose that for any clique C and any two vertices x,y\in C, there exists a wall in \mathcal{W}(C) separating x and y. If two hyperplanewalls M_{1}\in\mathcal{W}(J_{1}) and M_{2}\in\mathcal{W}(J_{2}) are tangent then either J_{1} and J_{2} are tangent or J_{1}=J_{2} and there exist two tangent walls W_{1},W_{2}\in\mathcal{W}(C), where C is a clique dual to J_{1}=J_{2}, such that M_{1}=\overline{W}_{1} and M_{2}=\overline{W}_{2}.
Proof.
First, notice that J_{1} and J_{2} cannot be transverse, since otherwise M_{1} and M_{2} would be transverse as well according to Lemma 4.13. Therefore, either J_{1} and J_{2} are nested or J_{1}=J_{2}. If J_{1} and J_{2} are nested but not tangent, there exists a hyperplane J separating J_{1} and J_{2}, and taking a wall M\in\mathcal{W}(J) separating \mathrm{proj}_{N(J)}(N(J_{1})) and \mathrm{proj}_{N(J)}(N(J_{2})) produces a hyperplanewall separating M_{1} and M_{2}. Therefore, if J_{1} and J_{2} are nested, they must be tangent. Finally, suppose that J_{1}=J_{2}, so that there exist two walls W_{1},W_{2}\in\mathcal{W}(C), where C is a clique dual to J_{1}=J_{2}, such that M_{1}=\overline{W}_{1} and M_{2}=\overline{W}_{2}. Clearly, if there exists a wall W\in\mathcal{W}(C) separating W_{1} and W_{2}, then \overline{W} separates M_{1} and M_{2}, so W_{1} and W_{2} must be tangent. ∎
For instance, setting \mathcal{W}(C)=\{\{\{x\},\{x\}^{c}\}\mid x\in C\} for every clique C produces a coherent system of wallspaces. Notice that, if C has cardinality two, then we introduced two indistinguishable walls; for convenience, we identify these two walls. We denote by \mathcal{SW} the resulting collection of walls on X, which we refer to as sectorwalls. Notice that a sectorwall is the data of a sector and its corresponding cosector (as defined in Section 2.8). This provides a systematic way to cubulate quasimedian graphs, allowing us to show that admitting a “nice” action on a quasimedian graph is equivalent to admitting a “nice” action on a CAT(0) cube complex. This idea is made precise by the following proposition.
Proposition 4.16.
Let G be a group acting on a quasimedian graph X. Let C(X,\mathcal{SW}) denote the CAT(0) cube complex obtained by (quasi)cubulating (X,\mathcal{SW}). Suppose that one of the following assertions holds:

•
G\curvearrowright X is not elliptic;

•
G\curvearrowright X is metrically proper;

•
G\curvearrowright X is geometric.
Then the action G\curvearrowright C(X,\mathcal{SW}) satisfies the corresponding property.
Essentially, Proposition 4.16 will be a consequence of the various observations made by the next preliminary lemmas.
Lemma 4.17.
For every vertices x,y\in X, d(x,y)\leq d_{\mathcal{SW}}(x,y)\leq 2d(x,y).
Proof.
By definition of \mathcal{SW}, d_{\mathcal{SW}}(x,y) is equal to the number of hyperplanes separating x and y counted with multiplicity, depending on whether it delimits more than two sectors. More precisely, if we set
n(J)=\left\{\begin{array}[]{cl}1&\text{if $J$ delimits two sectors}\\ 2&\text{if $J$ delimits at least three sectors}\end{array}\right. 
and if J_{1},\ldots,J_{n} are the hyperplanes separating x and y, then
d_{\mathcal{SW}}(x,y)=\sum\limits_{i=1}^{n}n(J_{i}). 
Since 1\leq n(J_{i})\leq 2 for every 1\leq i\leq n, and n=d(x,y), the conclusion follows. ∎
Corollary 4.18.
Let X be a quasimedian graph and let C(X,\mathcal{SW}) denote the CAT(0) cube complex obtained by (quasi)cubulating (X,\mathcal{SW}). For any H\leq\mathrm{Aut}(X), the induced action H\curvearrowright C(X,\mathcal{SW}) is elliptic if and only if H stabilises a prism of X.
Proof.
Let x\in X be a vertex, and let \sigma_{x} denote the associated principal orientation, thought of as a vertex of C(X,\mathcal{SW}). Noticing that, for every g\in H,
d_{C(X,\mathcal{SW})}(\sigma_{x},g\cdot\sigma_{x})=d_{C(X,\mathcal{SW})}(% \sigma_{x},\sigma_{g\cdot x})=d_{\mathcal{SW}}(x,g\cdot x), 
we deduce from Lemma 4.17 that the action H\curvearrowright C(X,\mathcal{SW}) has a bounded orbit if and only if the action H\curvearrowright X has a bounded orbit as well. On the other hand, we know from Theorem 2.115 that H\curvearrowright X has a bounded orbit if and only if H stabilises a prism. ∎
Lemma 4.19.
Two sectorwalls are transverse if and only if their underlying hyperplanes are transverse.
Proof.
For every clique and every vertices x,y\in C, the walls
\{\{x\},\{x\}^{c}\},\ \{\{y\},\{y\}^{c}\}\in\mathcal{W}(C) 
are nested, so that our lemma follows from Lemma 4.13. ∎
For our next and last preliminary lemma, we need the following definition:
Definition 4.20.
The simplicial dimension of a quasimedian graph X, denoted by \dim_{\triangle}X, is the maximal cardinality of a clique of X. The dimension of X is \dim(X)=\max(\dim_{\square}X,\dim_{\triangle}X).
Lemma 4.21.
Let X be a quasimedian graph of finite dimension and let C(X,\mathcal{SW}) denote the CAT(0) cube complex obtained by (quasi)cubulating (X,\mathcal{SW}). There exists an \mathrm{Aut}(X)invariant surjective map from the maximal cubes of C(X,\mathcal{SW}) to the maximal prisms of X such that the preimage of a maximal prism of C(X,\mathcal{SW}) has cardinality at most \dim_{\square}(X)\cdot\dim_{\triangle}(X).
Proof.
Let C be a maximal cube of C(X,\mathcal{SW}). Because \dim C(X,\mathcal{SW})=\dim_{\square}X<+\infty, this cube is uniquely determined by the hyperplanes dual to it, which produces a maximal collection of pairwise transverse hyperplanes (see Proposition 2.79). According to Theorem 2.56, to such a collection corresponds a maximal collection of pairwise transverse sectorwalls of X, producing a maximal collection of pairwise transverse hyperplanes of X according to Lemma 4.19, and finally a maximal prism P(C) according to Proposition 2.79. Notice that, according to the various results we used, the map C\mapsto P(C) is \mathrm{Aut}(X)invariant.
By construction, two maximal cubes C,C^{\prime} of C(X,\mathcal{SW}) satisfy P(C)=P(C^{\prime}) if and only if the two collections of pairwise transverse sectorwalls produce the same collection of pairwise transverse hyperplanes. Because there exist at most \dim_{\square}(X) pairwise transverse hyperplanes and that each hyperplane delimits at most \dim_{\triangle}(X) sectors, we deduce that the cardinality of a preimage of C\mapsto P(C) is at most \dim_{\square}(X)\cdot\dim_{\triangle}(X). ∎
Proof of Proposition 4.16..
The first point is a direct consequence of Corollary 4.18, and the second one a direct consequence of Lemma 4.17. Now, suppose that G acts geometrically on X. We already know that the action G\curvearrowright C(X,\mathcal{SW}) is metrically proper according to the previous point. Next, since a group acts geometrically on X, necessarily X must be uniformly locally finite. In particular, X is finite dimensional, so that Lemma 4.21 applies. Because the action G\curvearrowright X has finitely many orbits of maximal prisms, we deduce that the action G\curvearrowright C(X,\mathcal{SW}) has finitely many orbits of maximal cubes. Consequently, G\curvearrowright C(X,\mathcal{SW}) is geometric. ∎
4.2 Actions of groups
In this section, we are interested in group actions. First notice that, if a group G acts on a quasimedian graph X, endowed with a system of wallspaces which is coherent and Ginvariant (ie., g\mathcal{W}(C)=\mathcal{W}(gC) for every clique C and every g\in G), then G acts on the space with walls (X,\mathcal{HW}). Now, we would like to determine when this induced action G\curvearrowright(X,\mathcal{HW}) is metrically proper, cocompact, virtually special, etc. We give some criteria below.
Proposition 4.22.
Let G be a group acting on a quasimedian graph X endowed with a Ginvariant coherent system of wallspaces. Suppose that

•
any vertex of X belongs to finitely many cliques;

•
any vertexstabiliser is finite;

•
for every clique C, any two points of C are separated by a wall of \mathcal{W}(C);

•
for every clique C, the space with walls (C,\mathcal{W}(C)) is locally finite.
Then the action G\curvearrowright(X,\mathcal{HW}) is metrically proper.
Proof.
Let us consider the system of pseudometrics (C,d_{\mathcal{W}(C)}). Notice that, because any two points of a clique are sepated by a wall, this is a system of metrics. As a consequence of Fact 4.12, this system is coherent, and according to Lemma 4.11, d_{\mathcal{HW}} is the global metric extending this system. Thus, our proposition is a direct consequence of Lemma 3.24. ∎
For the next proposition, we need to introduce some notation. If P=C_{1}\times\cdots C_{n} is a prism of X, we denote by \mathcal{W}(P) the collection of walls \mathcal{W}(C_{1})\times\cdots\times\mathcal{W}(C_{n}).
Proposition 4.23.
Let G be a group acting on a quasimedian graph X endowed with a Ginvariant coherent system of wallspaces. Suppose that

•
the cubical dimension of X is finite;

•
for every clique C, \mathcal{W}(C)\neq\emptyset and \dim(C,\mathcal{W}(C))<+\infty;

•
X contains finitely many Gorbits of prisms;

•
for every maximal prism P, the action \mathrm{stab}(P)\curvearrowright(P,\mathcal{W}(P)) is cocompact.
Then the action G\curvearrowright(X,\mathcal{HW}) is cocompact.
Proof.
Define an Mcollection as the data of a maximal prism P and, for each hyperplane J dual to it, a maximal collection of pairwise transverse walls of \mathcal{W}(J)\subset\mathcal{W}(P). We claim that any maximal collection of pairwise transverse hyperplanewalls naturally defines an Mcollection. It is worth noticing that \dim(X,\mathcal{HW}) is finite according to Corollary 4.14, so that such a collection must be finite.
Let M_{1},\ldots,M_{n}\in\mathcal{HW} be a maximal collection of pairwise transverse hyperplanewalls. Let J_{1},\ldots,J_{m} denote the associated collection of pairwise distinct hyperplanes of X; according to Lemma 4.13, these hyperplanes are pairwise transverse. If there exists a hyperplane J^{\prime} transverse to J_{1},\ldots,J_{m}, then, according to Lemma 4.13, any wall M^{\prime}\in\mathcal{W}(J^{\prime}) (there exists such a wall since \mathcal{W}(J^{\prime}) is non empty by assumption) will be transverse to M_{1},\ldots,M_{n}, contradicting the maximality of our collection. Therefore, J_{1},\ldots,J_{m} is a maximal collection of pairwise transverse hyperplanes. Let P denote the maximal prism which is associated to this collection by Proposition 2.79. Fix some 1\leq j\leq n, and let M_{i_{1}},\ldots,M_{i_{k}} be the hyperplanewalls of our collection with J_{j} as underlying hyperplane. If there existed a hyperplanewall M\in\mathcal{W}(J_{j}) transverse to M_{i_{1}},\ldots,M_{i_{k}} then, according to Lemma 4.13, M would be transverse to M_{1},\ldots,M_{n}, contradicting the maximality of our collection. Therefore, M_{i_{1}},\ldots,M_{i_{k}} is a maximal collection of pairwise transverse hyperplanewalls in \mathcal{W}(J_{j}). Thus, P and M_{1},\ldots,M_{n} naturally defines an Mcollection.
On the other hand, we know that there exist finitely many Gorbits of (maximal) prisms of X and, for each maximal prism P, the action \mathrm{stab}(P)\curvearrowright(P,\mathcal{W}(P)) is cocompact, so there must exist only finitely many Gorbits of Mcollections. A fortiori, there exist only finitely many Gorbits of maximal collections of pairwise transverse hyperplanewalls in \mathcal{HW}, ie., the action G\curvearrowright(X,\mathcal{HW}) is cocompact. ∎
Before stating our next proposition, recall that a subgroup H\leq G is a retract if there exists an epimorphism r:G\twoheadrightarrow H (called a retraction) satisfying r_{H}=\mathrm{Id}_{H}.
Proposition 4.24.
Let G be a group acting on a quasimedian graph X endowed with a Ginvariant coherent system of wallspaces. Suppose that

•
for every clique C, two distinct vertices of C are separated by some wall of \mathcal{W}(C);

•
the action G\curvearrowright X is special.
Then \mathrm{Obs}(G\curvearrowright(X,\mathcal{HW}))=\bigcup\limits_{J\ \mathrm{% hyperplane}}\mathrm{Obs}(\mathrm{stab}(J)\curvearrowright(N(J),\mathcal{W}(J))). If moreover

•
for every hyperplane J, \mathrm{stab}(J) is a finitelygenerated retract of G;

•
X has finitely many Gorbits of hyperplanes;

•
for every hyperplane J, \mathrm{stab}(J)\curvearrowright(N(J),\mathcal{W}(J)) is virtually special;
then G\curvearrowright(X,\mathcal{HW}) is virtually special.
We recall that \mathcal{W}(J) denotes the collection of extended walls \overline{\mathcal{W}}(C), where C is any clique dual to J. (This collection does not depend on the choice of C since our system is coherent.)
Proof of Proposition 4.24..
Let g\in\mathrm{Obs}(G\curvearrowright(X,\mathcal{HW})). For convenience, let \mathrm{Obs}(J) denote the obstruction \mathrm{Obs}(\mathrm{stab}(J)\curvearrowright(N(J),\mathcal{W}(J))) for every hyperplane J of X. Three cases may happen.
Suppose that there exists a wall M\in\mathcal{HW} such that M and gM are transverse. Let J denote the underlying hyperplane of M. According to Lemma 4.13, either gJ=J or J and gJ are transverse. The latter case is impossible since G\curvearrowright X is special. Therefore, g\in\mathrm{stab}(J). We conclude that g\in\mathrm{Obs}(J).
Suppose that there exists a wall M\in\mathcal{HW} such that M and gM are tangent. Let J denote the underlying hyperplane of M. Notice that J and gJ cannot be tangent since G\curvearrowright X is special. Therefore, we deduce from Lemma 4.15 that g\in\mathrm{Obs}(J).
Suppose finally that there exist three walls M_{1},M_{2},M_{3}\in\mathcal{HW} such that M_{1} and M_{2} are tangent, M_{2} and M_{3} are transverse, and gM_{1}=M_{3}. Let J_{1},J_{2},J_{3} denote the underlying hyperplanes of M_{1},M_{2},M_{3} respectively; notice that gJ_{1}=J_{3}. We distinguish five cases.

•
If J_{1},J_{2},J_{3} are pairwise distinct, then J_{1} and J_{2} must be tangent and J_{2} and J_{3} transverse, which is impossible since G\curvearrowright X is special.

•
If J_{1}=J_{2}\neq J_{3}, then gJ_{2}=gJ_{1}=J_{3} is transverse to J_{2}, contradicting the fact that G\curvearrowright X is special.

•
If J_{2}=J_{3}\neq J_{1}, then g^{1}J_{2}=g^{1}J_{3}=J_{1} is tangent to J_{2}, contradicting the fact that G\curvearrowright X is special.

•
If J_{1}=J_{3}\neq J_{2}, then J_{2} is both transverse and tangent to J_{1}=J_{3}, which is impossible.

•
If J_{1}=J_{2}=J_{3}, let J denote this common hyperplane. In particular, gJ=gJ_{1}=J_{3}=J hence g\in\mathrm{stab}(J). We conclude that g\in\mathrm{Obs}(J).
We have proved that \mathrm{Obs}(G\curvearrowright(X,\mathcal{HW}))\subset\bigcup\limits_{J\ % \mathrm{hyperplane}}\mathrm{Obs}(\mathrm{stab}(J)\curvearrowright(N(J),% \mathcal{W}(J))). The reverse inclusion is clear, concluding the proof of the first assertion of our proposition.
Now, suppose that hyperplanestabilisers are finitely generated retracts, that X has finitely many Gorbits of hyperplanes, and that the actions \mathrm{stab}(J)\curvearrowright(N(J),\mathcal{W}(J)) are all virtually special.
Let J_{1},\ldots,J_{n} be a set of representatives of the hyperplanes of X under the action of G. Let 1\leq i\leq n. Because the action \mathrm{stab}(J_{i})\curvearrowright(N(J_{i}),\mathcal{W}_{i}) is virtually special, and that \mathrm{stab}(J_{i}) is finitely generated, there exists a normal finiteindex subgroup H_{i}\leq\mathrm{stab}(J_{i}) such that the induced action H_{i}\curvearrowright(N(J_{i}),\mathcal{W}(J_{i})) is special. In particular, if F_{i} denotes a set of representatives of \mathrm{stab}(J_{i})/H_{i} different from H_{i}, then
\mathrm{Obs}(\mathrm{stab}(J_{i})\curvearrowright(N(J_{i}),\mathcal{W}(J_{i}))% )\subset F_{i}H_{i} 
according to Fact 4.8. Fixing a retraction r_{i}:G\to\mathrm{stab}(J_{i}), we introduce
K_{i}=\mathrm{ker}\left(G\overset{r_{i}}{\twoheadrightarrow}\mathrm{stab}(J_{i% })\twoheadrightarrow\mathrm{stab}(J_{i})/H_{i}\right). 
Notice that K_{i} is a normal finiteindex subgroup of G satisfying H_{i}\leq K_{i} and K_{i}\cap F_{i}=\emptyset. Set K=\bigcap\limits_{i=1}^{n}K_{i}. This defines a normal finiteindex subgroup of G.
We claim that the induced action K\curvearrowright(X,\mathcal{HW}) is special. Let
g\in\mathrm{Obs}(G\curvearrowright(X,\mathcal{HW}))\subset\bigcup\limits_{k\in G% }\bigcup\limits_{i=1}^{n}kF_{i}H_{i}k^{1}, 
say g=kfhk^{1} for some 1\leq i\leq n, k\in G, f\in F_{i} and h\in H_{i}. Then we deduce from g\in K that
f\in k^{1}Kkh^{1}=Kh^{1}\subset K_{i}h^{1}=K_{i}, 
contradicting the fact that K_{i}\cap F_{i}=\emptyset. Therefore, K\cap\mathrm{Obs}(G\curvearrowright(X,\mathcal{HW}))=\emptyset. ∎
Proposition 4.25.
Let G be a group acting on a quasimedian graph X endowed with a Ginvariant coherent system of wallspaces. Suppose that

•
for every clique C, any two vertices of C are separated by a wall of \mathcal{W}(C);

•
for every prism P of X, the action \mathrm{stab}(P)\curvearrowright(P,\mathcal{W}(P)) is properly discontinuous;

•
vertexstabilisers are finite.
Then the action G\curvearrowright(X,\mathcal{HW}) is properly discontinuous.
Recall that the action G\curvearrowright(X,\mathcal{HW}) is properly discontinuous if the action of G on the CAT(0) cube complex obtained by cubulating (X,\mathcal{HW}) is properly discontinuous. Since an action on a CAT(0) cube complex is properly discontinuous if and only if its vertexstabilisers are finite, Proposition 4.25 precisely means that the stabiliser of any orientation of \mathcal{HW} must be finite. Of course, the first step toward the proof of our proposition is to understand the orientations of \mathcal{HW}.
Fix a prism P of X which is a Cartesian product of cliques C_{1}\times\cdots\times C_{n}, and, for every 1\leq i\leq n, let \sigma_{i} be a non principal orientation of \mathcal{W}(C_{i}). In particular, \sigma_{P}=\prod\limits_{i=1}^{n}\sigma_{i} defines an orientation of \mathcal{W}(P). Now, we extend \sigma_{P} on \mathcal{HW} in the following way: if W\in\mathcal{W}(J) for some hyperplane J which is disjoint from P, set \sigma_{P}(W) as the unique halfspace of W containing P. It is not difficult to verify that \sigma_{P} defines an orientation. An orientation arising in this way will be called semiprincipal.
Lemma 4.26.
Let X be a quasimedian graph endowed with a coherent system of wallspaces. Suppose that, for every clique C of X, any two vertices of C are separated by a wall of \mathcal{W}(C). Then any orientation of \mathcal{HW} is either principal or semiprincipal.
Proof.
Fix some orientation \sigma of \mathcal{HW}. Let \sigma_{x} be a principal orientation minimizing the distance to \sigma in C(X,\mathcal{HW}), ie., the number of walls on which \sigma_{x} and \sigma differ, and let \mathfrak{I} denote the collection of walls on which they differ.
We claim that, for any hyperplane J associated to a wall W\in\mathfrak{I}, we have x\in N(J). So fix a wall W\in\mathfrak{I} and let J denote its underlying hyperplane. Let y denote the projection of x onto N(J). Of course, \sigma_{x}(W^{\prime})=\sigma_{y}(W^{\prime}) for any wall W^{\prime}\in\mathcal{HW} which does not separate x and y. Now, suppose that W^{\prime}\in\mathcal{HW} is a wall separating x and y. In particular, its underlying hyperplane J^{\prime} separates x and y. As a consequence of Lemma 2.34, J^{\prime} separates x and N(J), so that N(J) must be included into a single sector S delimited by J^{\prime}; notice that y\in S. On the other hand, because \sigma_{x}(W)\neq\sigma(W), we know that \sigma(W) does not contain x, so \sigma(W)\subset S. By noticing that \sigma(W)\subset S\subset\sigma_{y}(W^{\prime}), we conclude that
\sigma(W^{\prime})=\sigma_{y}(W^{\prime})\neq\sigma_{x}(W^{\prime}). 
Therefore, since \sigma_{x} is a principal orientation minimizing the number of walls on which \sigma_{x} and \sigma differ, we deduce that there exist no wall of \mathcal{HW} separating x and y. On the other hand, a hyperplane H separating x and y would produce a wall separating x and y by taking a wall of \mathcal{W}(H) separating the projections of x and y onto N(H) (such a wall would exist by assumption), so we deduce that no hyperplane separates x and y, ie., x=y. A fortiori, x\in N(J).
We claim that any two hyperplanes underlying two walls of \mathfrak{I} are transverse. Let J_{1},J_{2} be two non transverse hyperplanes such that x\in N(J_{1})\cap N(J_{2}) and such that J_{1} is the underlying hyperplane of some W_{1}\in\mathfrak{I}. Fix a wall W_{2}\in\mathcal{W}(J_{2}). Because W_{1}\in\mathfrak{I}, we know that \sigma(W_{1})\neq\sigma_{x}(W_{1}), hence \sigma(W_{1})\subset\sigma_{x}(W_{2}). This implies that \sigma(W_{2})=\sigma_{x}(W_{2}); a fortiori, W_{2}\notin\mathfrak{I}. This proves our claim.
Thus, if \mathfrak{H} denotes the set of the hyperplanes underlying some wall of \mathfrak{I}, we have proved that \mathfrak{H} defines a (finite) collection of pairwise transverse hyperplanes satisfying x\in\bigcap\limits_{J\in\mathfrak{H}}N(J). It follows from Fact 2.75 that there exists a prism P=C_{1}\times\cdots\times C_{n} such that x\in P and such that \mathfrak{H} is precisely the set of the hyperplanes dual to it.
Now, we claim that, for every 1\leq i\leq n, the restriction \sigma_{\mathcal{W}(C_{i})} defines an orientation of \mathcal{W}(C_{i}) which is not principal. Indeed, suppose by contradiction that there exists some 1\leq i\leq n such that the restriction \sigma_{\mathcal{W}(C_{i})} is a principal orientation of \mathcal{W}(C_{i}); let y\in C_{i} denote the corresponding vertex. Because x and y both belong to C_{i}, the orientations \sigma_{x} and \sigma_{y} may only differ on walls with underlying hyperplane J_{i}, the hyperplane dual to C_{i}. On the other hand, \sigma_{y} and \sigma coincide on every wall with underlying hyperplane J_{i}. It follows that the number of walls on which \sigma_{y} and \sigma differ is stricly smaller than the number of walls on which \sigma_{x} and \sigma differ, contradicting our choice of \sigma_{x}. This proves our claim.
Let \sigma_{P} denote the semiprincipal orientation associated to \prod\limits_{i=1}^{n}\sigma_{\mathcal{W}(C_{i})}. We claim that \sigma=\sigma_{P}. Let W\in\mathcal{HW} be a wall and let J denote its underlying hyperplane. If W\in\mathcal{W}(C_{i}) for some 1\leq i\leq n (ie., if J intersects the prism P), then by construction we know that \sigma(W)=\sigma_{P}(W). Next, if J is disjoint from P, \sigma_{P}(W) must be the halfspace delimited by W which contains P (or equivalently, x\in P), hence
\sigma_{P}(W)=\sigma_{x}(W)=\sigma(W) 
because J\notin\mathfrak{H} so that W\notin\mathfrak{I}. A fortiori, \sigma is a semiprincipal orientation. ∎
Proof of Proposition 4.25..
In order to prove that the action G\curvearrowright(X,\mathcal{HW}) is properly discontinuous, we will prove that the stabiliser of any orientation \sigma of \mathcal{HW} is finite. According to Lemma 4.26, we know that \sigma is either principal or semiprincipal. In the former case, say \sigma=\sigma_{x} for some vertex x\in X, we have \mathrm{stab}(\sigma_{x})=\mathrm{stab}(x), which is finite by assumption. Next, suppose that \sigma is semiprincipal. So there exists a prism P=C_{1}\times\cdots\times C_{n} such that, for every 1\leq i\leq n, the restriction \sigma_{\mathcal{W}(C_{i})} defines an orientation which is not principal, and such that, for any wall W\in\mathcal{HW} whose underlying hyperplane does not intersect P, \sigma(W) is the halfspace delimited by W which contain P. Let \mathfrak{H} denote the set of the hyperplanes J such that the restriction \sigma_{\mathcal{W}(J)} is principal, and let S(J) denote the sector delimited by J which contains P; notice that, if we fix some clique C dual to J, then S(J) is also the sector [C,x] where x\in C is defined by \sigma_{\mathcal{W}(C)}=\sigma_{x}. The intersection C=\bigcap\limits_{J\in\mathfrak{H}}S(J) must be \mathrm{stab}(\sigma)invariant since the collection \mathfrak{H} and the sectors S(J) are defined only from \sigma. We claim that C=P.
Let J\in\mathfrak{H}. Fix a clique C dual to J and let x\in C be the vertex defined by \sigma_{\mathcal{W}(C)}=\sigma_{x}, so that S(J)=[C,x]. Suppose by contradiction that x is different from the projection y of P onto C. Then there exists a wall W\in\mathcal{W}(C) separating x and y. Clearly, \sigma(W)=\sigma_{x}(W) is disjoint from P since y\notin\sigma(W). On the other hand, by definition of P, we know that P\subset\sigma(W^{\prime}) for every W^{\prime}\in\mathcal{W}(C) since J does not intersect P, whence a contradiction. Therefore, x=y holds, and we deduce that S(J) is the sector delimited by J containing P. A fortiori, P\subset C. Conversely, let x\in X be a vertex satisfying x\notin P. Let y\in P denote the projection of x onto P (which is welldefined since prisms are gated according to Lemma 2.80). Because x\notin P, necessarily x\neq y so that there exists a hyperplane J separating x and y. Notice that J does not intersect P according to Lemma 2.34, so J\in\mathfrak{H}. Moreover, we saw that S(J) is the sector delimited by J which contains P, so x\notin S(J) and C\subset S(J), hence x\notin C. A fortiori, C\subset P, which concludes our claim.
Thus, we have proved that the prism P is \mathrm{stab}(\sigma)invariant. On the other hand, we know by assumption that the action \mathrm{stab}(P)\curvearrowright(P,\mathcal{W}(P)) is properly discontinuous, so that the stabiliser (in \mathrm{stab}(P)) of the orientation \prod\limits_{i=1}^{n}\sigma_{\mathcal{W}(C_{i})} of \mathcal{W}(P) is finite. We conclude that the stabiliser (in G) of \sigma must be finite as well. ∎
4.3 From actions to walls
In order to apply the different propositions proved in the previous section, we need to find a collection of walls on each clique of our quasimedian graph. In the cases we will be interested in, the induced action of each cliquestabiliser on the corresponding clique often turns out to be transitive and free, so that finding walls on a clique is equivalent to finding walls on its stabiliser. For this purpose, one possibility is to consider an action on a CAT(0) cube complex and then to pullback the walls defined on the cube complex to our group. In fact, the construction presented below works in full generality for actions on quasimedian graphs, but, as noticed by Proposition 4.16, we do not really loss generality.
Let G be a group acting on a CAT(0) cube complex X. Fix a base vertex x_{0}\in X. Without loss of generality, we suppose that the combinatorial convex hull of the orbit G\cdot x_{0} is the whole X. To any hyperplane J of X, delimiting two halfspaces J^{+} and J^{}, we associate the wall on G
\mathfrak{m}(J)=\left\{\{g\in G\mid g\cdot x_{0}\in J^{+}\},\{g\in G\mid g% \cdot x_{0}\in J^{}\}\right\}; 
and we introduce the collection of walls
\mathcal{M}(G\curvearrowright X)=\left\{\mathfrak{m}(J)\mid J\ \text{% hyperplane of}\ X\right\}. 
Notice that, for every hyperplane J and every g\in G, we have g\cdot\mathcal{M}(J)=\mathcal{M}(g\cdot J), so that \mathcal{M}(G\curvearrowright X) defines a Ginvariant collection of walls. Moreover, for every hyperplane J of X and every g,h\in G, the wall \mathfrak{m}(J) separates g and h if and only if J separates g\cdot x_{0} and h\cdot x_{0}, so that
d_{\mathcal{M}(G\curvearrowright X)}(g,h)=d_{X}(g\cdot x_{0},h\cdot x_{0}). 
As a consequence, (G,\mathcal{M}=\mathcal{M}(G\curvearrowright X)) defines a space with walls. But does the properties of the action G\curvearrowright X transfer to G\curvearrowright(G,\mathcal{M})? The next two lemmas provide positive answers for the metrical properness and the cocompactness.
Lemma 4.27.
If G\curvearrowright X is metrically proper, then so is G\curvearrowright(G,\mathcal{M}).
Proof.
Because the action G\curvearrowright X is metrically proper, we deduce that for every R\geq 0
\#\{g\in G\mid d_{\mathcal{M}}(1,g)\leq R\}=\#\{g\in G\mid d_{X}(x_{0},g\cdot x% _{0})\leq R\}<+\infty. 
Therefore, the action G\curvearrowright(G,\mathcal{W}) is metrically proper as well. ∎
Lemma 4.28.
If G\curvearrowright X is cocompact and X locally finite, then G\curvearrowright(G,\mathcal{M}) is cocompact.
Proof.
Let Y\subset X be a compact fundamental domain containing x_{0}. Because X is locally finite, the 1neighborhood Y^{+1} of Y must be finite. Let J_{1},\ldots,J_{p} denote the hyperplanes intersecting Y^{+1}. For every 1\leq i\leq p, there exists some g_{i}\in G such that J_{i} separates x_{0} and g_{i}\cdot x_{0}. Set R=\max\{d(x_{0},g_{i}\cdot x_{0})\mid 1\leq i\leq p\}.
Let \mathfrak{m}(H_{1}),\ldots,\mathfrak{m}(H_{n})\in\mathcal{M} be a collection of pairwise transverse walls. Up to reordering these walls, suppose that, for some 1\leq s\leq n, H_{1},\ldots,H_{s} is a maximal subcollection of H_{1},\ldots,H_{n} of pairwise transverse hyperplanes. If we fix some x\in\bigcap\limits_{i=1}^{s}N(H_{i}), we can suppose, up to a translation, that x\in Y, so that H_{1},\ldots,H_{s}\in\{J_{1},\ldots,J_{p}\}.
Now, fix some s+1\leq j\leq n. We know that H_{j} must be disjoint from some H_{k}, 1\leq k\leq n. Let H_{j}^{+} denote the halfspace delimited by H_{j} which contains H_{k}, and similarly, let H_{k}^{+} denote the halfspace delimited by H_{k} which contains H_{j}. If G\cdot x_{0} intersected H_{j}^{+}\cap H_{k}^{+}, then \mathfrak{m}(H_{j}) and \mathfrak{m}(H_{k}) would be nested. So, because we supposed \mathfrak{m}(H_{j}) and \mathfrak{m}(H_{k}) transverse, necessarily G\cdot x_{0} does not intersect H_{j}^{+}\cap H_{k}^{+}, so that H_{j} and H_{k} are indistinguishable. As a consequence, there exists some 1\leq\ell\leq p such that H_{j} and J_{\ell}=H_{k} are indistinguishable. In particular, H_{j} must separate x_{0} and g_{\ell}\cdot x_{0}, so that H_{j} intersects the ball B(x_{0},R)\subset Y^{+R}.
Thus, we have proved that H_{i} intersects Y^{+R} for every 1\leq i\leq n. Notice that, because X is locally finite, Y^{+R} must be finite. In particular, the cardinality n of our collection is bounded above by the (finite) number of hyperplanes intersecting Y^{+R}, hence
Fact 4.29.
If G\curvearrowright X is cocompact and X locally finite, then \dim(G,\mathcal{M})<+\infty.
But we have also proved that there exist only finitely many Gorbits of (maximal) collections of pairwise transverse walls of \mathcal{M}, ie., G\curvearrowright(G,\mathcal{M}) is cocompact. ∎
Corollary 4.30.
If G\curvearrowright X is geometric, then so is G\curvearrowright(G,\mathcal{M}).
It is worth noticing that the dimension of (G,\mathcal{M}) does not depend only on X. To be precise, if given a hyperplane J of X we denote by \iota(J) the number of hyperplanes indistinguishable to J, we have
\dim(G,\mathcal{M})=\max\left\{\sum\limits_{i=1}^{n}\iota(J_{i})\mid J_{1},% \ldots,J_{n}\ \text{pairwise transverse}\right\}. 
Nevertheless, we know that the dimension is finite whenever the cube complex X is finite dimensional. In fact, for geometric actions, it is possible to introduce another space with walls with a better control on the dimension. If G acts on a CAT(0) cube complex X, we denote by \mathcal{N}=\mathcal{N}(G\curvearrowright X) the collection of walls obtained from \mathcal{M} by identifying two indistinguishable walls. Explicitely, the walls of \mathcal{N} are the \mathfrak{m}(J)’s where \mathfrak{m}(J_{1})=\mathfrak{m}(J_{2}) if J_{1} and J_{2} are indistinguishable.
Lemma 4.31.
If the action G\curvearrowright X is geometric, then so is G\curvearrowright(G,\mathcal{N}). Moreover, \dim(G,\mathcal{N})\leq\dim X.
Proof.
First, we claim that there exists a constant R\geq 0 such that any two indistinguishable hyperplanes J_{1},J_{2} satisfies d(N(J_{1}),N(J_{2}))\leq R.
Let \{H_{1},\ldots,H_{n}\} be a set of representatives for the action of G on the hyperplanes of X. For every 1\leq i\leq n, fix some g_{i}\in G such that H_{i} separates x_{0} and g_{i}\cdot x_{0}. Set R=\max\{d(x_{0},g_{i}\cdot x_{0})\mid 1\leq i\leq n\}. Now, let J_{1},J_{2} be two indistinguishable hyperplanes. There exists some g\in G such that g\cdot J_{1}\in\{H_{1},\ldots,H_{n}\}, say g\cdot J_{1}=H_{1}. Therefore, the hyperplanes H_{1} and g\cdot J_{2} are necessarily indistinguishable, so that they both separate x_{0} and g_{1}\cdot x_{0}, hence
d(N(J_{1}),N(J_{2}))=d(N(H_{1}),N(gJ_{2}))\leq d(x_{0},g_{1}\cdot x_{0})\leq R. 
This proves our claim.
Now, fixing some x\in X and K\geq\dim(X), we want to prove that the set
F=\{g\in G\mid d_{\mathcal{N}}(x,g\cdot x)\leq K\} 
is finite. Let g\in G satisfy d(x,g\cdot x)\geq\mathrm{Ram}(K(R+3)+1), where \mathrm{Ram}(\cdot) denotes the Ramsey number. If so, we know that x and g\cdot x are separated by at least \mathrm{Ram}(K(R+3)+1) hyperplanes, so that x and g\cdot x must be separated by at least K(R+3)+1 disjoint hyperplanes, say J_{0},\ldots,J_{K(R+3)}. Because, for every 0\leq i\leq K1, the hyperplanes J_{i(R+3)} and J_{(i+1)(R+3)} are separated by at least R+1 hyperplanes, we deduce from our previous observation that J_{i(R+3)} and J_{(i+1)(R+3)} cannot be indistinguishable. Thus, \mathfrak{m}(J_{0}),\mathfrak{m}(J_{R+3}),\ldots,\mathfrak{m}(J_{K(R+3)}) define K+1 pairwise distinct walls of \mathcal{N} separating x and g\cdot x, hence g\notin F. Therefore,
F\subset\{g\in G\mid d(x,g\cdot x)\leq\mathrm{Ram}(K(R+3)+1)\}, 
and we conclude that F is finite because the action G\curvearrowright X is metrically proper. A fortiori, the action G\curvearrowright(X,\mathcal{N}) is metrically proper as well.
Finally, we claim that the action G\curvearrowright(X,\mathcal{N}) is cocompact, ie., there exist only finitely many orbits of (maximal) collections of pairwise transverse walls of \mathcal{N}. Notice that if \mathfrak{m}(J_{1}) and \mathfrak{m}(J_{2}) are transverse (as two elements of \mathcal{N}) then J_{1} and J_{2} must be transverse. Indeed, if J_{1} and J_{2} are two disjoint hyperplanes of X, and if we denote by J_{1}^{+} (resp. J_{2}^{+}) the halfspace delimited by J_{1} (resp. J_{2}) containing J_{2} (resp. J_{1}), then either J_{1}^{+}\cap J_{2}^{+}\cap G\cdot x_{0}=\emptyset, so that J_{1} and J_{2} are indistinguishable (and \mathfrak{m}(J_{1})=\mathfrak{m}(J_{2}) in \mathcal{N}), or \mathfrak{m}(J_{1}) and \mathfrak{m}(J_{2}) are nested. Therefore, any collection of pairwise transverse walls of \mathcal{N} defines a collection of pairwise transverse hyperplanes of X. Because we already know that X contains only finitely many orbits of collections of pairwise transverse hyperplanes, the conclusion follows. ∎
Question 4.32.
If G\curvearrowright X is geometric and virtually special, is G\curvearrowright(G,\mathcal{N}) virtually special as well?
In order to apply the propositions proved in the previous section, we also want any two distinct elements of G to be separated by some wall of \mathcal{M} or \mathcal{N}. The following lemma gives a sufficient condition which implies this property, and moreover Lemma 4.34 below essentially states that we can always assume that this condition is satisfied.
Lemma 4.33.
If the stabiliser of x_{0} is trivial, then two distinct elements of G are separated by some \mathfrak{m}(J).
Proof.
Let g,h\in G be two distinct elements. Because the stabiliser of x_{0} is trivial, necessarily g\cdot x_{0}\neq h\cdot x_{0}. Thus, if J denotes a hyperplane separating g\cdot x_{0} and h\cdot x_{0}, then \mathfrak{m}(J) must separate g and h. ∎
Lemma 4.34.
Let G be a group acting on a CAT(0) cube complex X_{0}. Then G acts on a CAT(0) cube complex X containing X_{0} so that the action G\curvearrowright X_{0} extends to an action G\curvearrowright X and X contains a vertex whose stabiliser is trivial. Moreover, the action G\curvearrowright X is properly discontinuous (resp. metrically proper, cocompact) if and only if the action G\curvearrowright X_{0} is properly discontinuous (resp. metrically proper, cocompact) as well; also, if G is a finitely generated residually finite group and if G\curvearrowright X_{0} is properly discontinous and virtually special, then G\curvearrowright X is virtually special.
Proof.
Let x_{0}\in X_{0} be a base vertex and let \Omega denote its Gorbit. Let X be the cube complex constructed from X_{0} by adding one vertex (x,g) for every x\in\Omega and g\in\mathrm{stab}(x), and one edge between x and (x,g) for every x\in\Omega and g\in\mathrm{stab}(x). It is clear that X satisfies the Gromovlink condition and that it is simply connected, so it is a CAT(0) cube complex.
Now, we extend the action G\curvearrowright X_{0} to an action G\curvearrowright X. For every x\in\Omega, fix some h_{x}\in G such that h_{x}\cdot x_{0}=x. For every g,k\in G and x\in\Omega, define
g\cdot(x,k)=(gx,gkh_{x}h_{gx}^{1}); 
notice that
gkh_{x}h_{gx}^{1}\cdot gx=gkh_{x}\cdot x_{0}=gk\cdot x=g\cdot x, 
so that gkh_{x}h_{gx}^{1}\in\mathrm{stab}(gx). Moreover,
\begin{array}[]{lcl}g_{1}\cdot(g_{2}\cdot(x,k))&=&g_{1}\cdot(g_{2}x,g_{2}kh_{x% }h_{g_{2}x}^{1})\\ \\ &=&(g_{1}g_{2}x,g_{1}\cdot g_{2}kh_{x}h_{g_{2}x}^{1}\cdot h_{g_{2}x}h_{g_{1}g% _{2}x}^{1})\\ \\ &=&(g_{1}g_{2}x,g_{1}g_{2}kh_{x}h_{g_{1}g_{2}x}^{1})=g_{1}g_{2}\cdot(x,k)\end% {array} 
so we have defined a group action G\curvearrowright X, which extends G\curvearrowright X_{0} by construction.
Fixing some x\in\Omega, we claim that the vertex (x,1)\in X has trivial stabiliser. Indeed, if g\in G fixes (x,1), then (x,1)=g\cdot(x,1)=(gx,gh_{x}h_{gx}^{1}). As a consequence, gx=x, ie., g\in\mathrm{stab}(x), so that h_{gx}=h_{x}. Therefore, our relation becomes (x,1)=(x,g), hence g=1.
This proves the first assertion of our lemma. Next, it is clear that the action G\curvearrowright X is properly discontinuous (resp. metrically proper, cocompact) if and only if the action G\curvearrowright X_{0} is properly discontinuous (resp. metrically proper, cocompact) as well. Finally, suppose that G is a finitely generated residually finite group and that G\curvearrowright X_{0} is properly discontinous and virtually special. Let H\leq G be a finiteindex subgroup acting specially on X_{0}. Then
\mathrm{Obs}(H\curvearrowright X)\subset\bigcup\limits_{x\in\Omega}\mathrm{% stab}(x)=\bigcup\limits_{g\in G}g\cdot\mathrm{stab}(x_{0})\cdot g^{1}. 
Because G acts properly discontinuously on X_{0}, \mathrm{stab}(x) must be finite. Therefore, G contains a normal finiteindex K such that K\cap\mathrm{stab}(x_{0})=\{1\}. A fortiori, K\cap\mathrm{Obs}(H\curvearrowright X)=\emptyset, so that H\cap K defines a finiteindex subgroup of G acting specially on X. Consequently, the action G\curvearrowright X is virtually special. ∎
4.4 Generalizations of spaces with walls
In [CMV04], Chelix, Martin and Valette introduces spaces with measured walls, which may be thought of as “continuous spaces with walls”; they may also be compared to real trees generalizing simplicial trees.
Definition 4.35.
A space with measured walls (X,\mathcal{W},\mathcal{B},\mu) is the data of a collection of walls \mathcal{W} on a set X, a \sigmaalgebra \mathcal{B} on \mathcal{W}, and a measure \mu on (\mathcal{W},\mathcal{B}), so that, for every x,y\in X, the set \mathcal{W}(x\mid y) of the walls separating x and y belongs to \mathcal{B} and has finite measure.
In particular, the definition allows us to introduce the pseudodistance
d_{\mathcal{W}}:(x,y)\mapsto\mu\left(\mathcal{W}(x\mid y)\right). 
Spaces with measured walls were essentially introduced because of their relations with Kazhdan’s property (T) and aTmenability (also known as Haagerup’s property). Recall that a (discrete) group is aTmenable if it admits a proper action on a Hilbert space by affine isometries. In this article, we are not interested in Kazhdan’s property (T) since, according to Proposition 4.16, any group acting without global fixed point on a quasimedian graph acts without global fixed point on a CAT(0) cube complex, and so does not satisfy Kazhdan’s property (T) according to [NR98b]. The relations between aTmenability and spaces with measured walls were strenghtened by Chatterji, Druţu and Haglund in [CDH10, Theorem 1.3] whe