1 Introduction
The order dimension (also known as the DushnikMiller dimension) of a poset has been introduced by Dushnik and Miller [10]. It is the minimum number of linear extensions of such that is the intersection of these extensions. See [20] for a comprehensive study of this topic. Schnyder [19] studied the DushnikMiller dimension of the incidence posets of graphs. Some classes of graphs can be characterized by their DushnikMiller dimension which is the DushnikMiller dimension of their incidence poset. For example, path forests are the graphs of DushnikMiller dimension at most . Schnyder [19] obtained a celebrated combinatorial characterization of planar graphs: they are those of DushnikMiller dimension at most . The question of characterizing classes of graphs of larger dimension is open. Nevertheless there are some partial results. Bayer et al. [1] and Ossona de Mendez [18] showed that every simplicial complex of DushnikMiller dimension has a straight line embedding in which generalizes the result of Schnyder in a way. The reciprocal is false by considering which has a straight line embedding in while it has DushnikMiller dimension [15]. The class of DushnikMiller dimension at most graphs is rather rich. Extremal questions in this class of graphs have been studied: Felsner and Trotter [14] showed that these graphs can have a quadratic number of edges. Furthermore, in order to solve a question about conflict free coloring [12], Chen et al. [8] showed that most of the graphs of DushnikMiller dimension only have independent sets of size at most . This result also implies that there is no constant such that every graph of DushnikMiller dimension at most is colorable. Therefore, graphs of DushnikMiller dimension at most seem difficult to characterize. Nevertheless, it was conjectured in [17] (See also [11]) that the class of DushnikMiller dimension complexes is the class of TDDelaunay complexes which we will define in the next paragraph. The result holds for and , but we disprove it already for in this paper.
We now define the class of TDDelaunay complexes which finds its origins in spanners introduced by Chew and Drysdale [5]. Given points in the plane, a plane spanner is a subgraph of the complete graph on these points which is planar when joining adjacent points with segments. The stretch of a plane spanner is the maximum ratio of the distance in the graph between two vertices when using the Euclidean weight function and the Euclidean distance between these two points. Given points in the plane, the question raised by Chew and Drysdale is to find a plane spanner which minimizes the stretch. We define the stretch of a class of plane spanners as the maximum stretch of any of these graphs. Chew [6] found the first class of plane spanners. It consists in the class of Delaunay graphs which is a variant of the Delaunay graphs where the norm is replaced by the norm. Given a norm and points in the plane, we define their Delaunay graph according to this norm as follows. Given a point , we define its Voronoi cell as the points of the plane such that is among the nearest points of (according to the norm ) to . Two points are connected if and only if their Voronoi cells intersect. This is equivalent to the fact that there exists a disk containing both points but no other in its interior. Chew [6] conjectured that the class of Delaunay graphs (classical Delaunay graphs) has a finite stretch. This question initiated a series of papers about this topic which drops the upper bound from and to [3, 9, 21]. A variant of this problem asks for minimizing the maximum degree. Kanj et al. [16] proved that there exists plane spanners of maximum degree 4 and of stretch at most .
Just a few years after introducing the plane spanner problem, Chew [7] found a second class of plane spanners. It consists in TDDelaunay graphs, obtained using the socalled triangular distance (which is not a distance but which is a convex distance function). Given a compact convex shape and a point in the interior of , we define the convex distance function, also called Minkowski distance function, between two points and as the minimal scaling factor such that after rescaling by and translating it in the way to center it on , then it contains also . By taking the unit circle we get the Euclidean distance. By taking the unit square we get the distance. By taking an equilateral triangle we get what we call the triangular distance. Chew [7] showed that their stretch is at most making the class of TDDelaunay graphs the best plane spanners class until Xia [21] showed that the stretch of Delaunay graphs is strictly less than . This class is also used to obtain bounded degree plane spanners [16]. TDDelaunay graphs can be generalized to higher dimensions by taking the triangular distance in according to a regular simplex.
The second section is dedicated to the notion of DushnikMiller dimension applied to the inclusion poset of simplicial complexes. In the third section we define TDDelaunay complexes. In the fourth section, we introduce the notion of multiflows which will be useful to the main theorem. The fifth section contains our main contribution in the form of a simplicial complex whose inclusion poset has DushnikMiller dimension but that is not a TDDelaunay complex in . Finally, in the sixth section we prove that rectangular Delaunay complexes [8, 13] are TDDelaunay complexes in .
2 DushnikMiller dimension of simplicial complexes
Bonichon et al. [2] showed the following property of TDDelaunay graphs.
Theorem 1
A graph is planar if and only if is a subgraph of a TDDelaunay graph.
As a graph is planar if and only if is of DushnikMiller dimension at most , there is maybe a link between the notions of DushnikMiller dimension and TDDelaunay complexes. In [17] and [11] it was independently conjectured that DushnikMiller dimension at most complexes are exactly the subcomplexes of TDDelaunay complexes of .
We now need to define formally the notions used. First of all abstract simplicial complexes generalize the notion of graphs. An abstract simplicial complex with vertex set is a set of subsets of which is closed by inclusion (i.e. , ). An element of is called a face. A maximal element of according to the inclusion order is called a facet.
2.1 DushnikMiller dimension
The notion of DushnikMiller dimension of a poset has been introduced by Dushnik and Miller [10]. It is also known as the order dimension of a poset.
Definition 2
The DushnikMiller dimension of a poset is the minimum number such that is the intersection of linear extensions of . This means that there exists extensions of such that for every , if and only if for every . In particular if and are incomparable with respect to , then there exists and such that and .
The notion of DushnikMiller dimension can be applied to an abstract simplicial complex as follows.
Definition 3
Let be an abstract simplicial complex. The inclusion poset of is the poset . The DushnikMiller dimension of is the DushnikMiller dimension of the inclusion poset of .
We denote the DushnikMiller dimension of a simplicial complex . Low dimensions are well known. We have if and only if is a vertex. We have if and only if is a union of paths. There are complexes with arbitrarily high DushnikMiller dimension: for any integer , where denotes the complete graph on vertices. The following theorem shows that the topological notion of planarity can be understood as a combinatorial property thanks to the DushnikMiller dimension.
Theorem 4 (Schnyder [19])
A graph is planar if and only if .
A generalization of the notion of planarity for simplicial complexes is the notion of straight line embedding. As in the planar case, we do not want that two disjoint faces intersect. We recall that a simplex of is the convex hull of a set of affinely independent points. We denote , the convex hull of a set of points .
Definition 5
Let be a simplicial complex with vertex set . A straight line embedding of in is a mapping such that

, is a set of affinely independent points of ,

, .
The following theorem shows that the DushnikMiller dimension in higher dimensions also captures some geometrical properties.
Theorem 6 (Bayer et al. [1], and Ossona de Mendez [18])
Any simplicial complex such that
has a straight line embedding in .
For , this theorem states that if a simplicial complex has dimension at most then it is planar. Brightwell and Trotter [4] proved that the converse also holds (for )^{1}^{1}1Note that in a straight line embedding in every triangle is finite, and it is thus impossible to embed a spherical complex like a octahedron or any polyhedron with triangular faces.. For higher , the converse is false: take for example the complete graph which has a straight line embedding in (and therefore in for ) and which has DushnikMiller dimension [15].
2.2 Representations
Representations have been introduced by Schnyder [19] in order to prove Theorem 4. It is a tool for dealing with DushnikMiller dimension. Here, only vertices will be ordered while in the DushnikMiller dimension every face of the complex must be ordered.
Definition 7
Given a linear order on a set , an element , and a set , we say that dominates in , and we denote it , if for every . A representation on a set is a set of linear orders on . Given a representation , an element , and a set , we say that dominates (in ) if dominates in some order . We define as the set of subsets of such that every dominates .
Note that for any representation . An element is a vertex of if . Note that sometimes an element is not a vertex of . Actually, the definition of representation provided here is slightly different from the one in [19] and [18]. There, the authors ask for the intersection of the orders to be an antichain. With this property, every element of is a vertex of . Note that simply removing the elements of that are not vertices of yields a representation in the sense of [18, 19].
Proposition 8
For any representation on a set , is an abstract simplicial complex.
Proof. For any , let be any subset of , and let be any element of . Since , there exists such that for every . Particularly, for every . Thus and we have proven that is an abstract simplicial complex.
An example is the following representation on where each line corresponds to a linear order whose elements appear in increasing order from left to right:
The corresponding complex is given by the facets and . For example is not in as does not dominate in any order. The following theorem shows that representations and DushnikMiller dimension are equivalent notions.
Theorem 9 (Ossona de Mendez [18])
Let be a simplicial complex with vertex set . Then if and only if there exists a representation on such that is included in .
For the following proofs, will be a representation and will be a representation . The following lemmas are technical and will be useful for the proof of our main result, Theorem 22.
Lemma 10
Let be a representation on . Let and be two different vertices of (i.e. and ) such that and such that and are consecutive in the order . The representation obtained after the permutation of and in the order , denoted , is such that .
Proof. Without loss of generality, we suppose that and that . Let us first show that . Towards a contradiction, let us consider a face such that . There exists therefore a vertex which does not dominate in . As dominates in , and as for every , we thus have that and . This implies that , that . As and are not adjacent, we have that . Furthermore, only dominates in order of . We denote the maximum vertex of in order , then and:
As , there exists an element such that for every . Thus either , contradicting the fact that , or for every (in particular for because and are consecutive), contradicting the fact that .
We showed that , but as this operation is an involution, we have that . Thus .
Note that is such that for every edge the orders between its endpoints are preserved. In other words, for any we have that if and only if . Given a representation , an increasing path is a path in such that for every .
Lemma 11
Let be a representation on , let be a vertex of (i.e. ), and let be any order of . There exists a representation such that:

,

for every , a path is increasing (in ) if and only if it is increasing (in ),

for every ,

if and only if there exists an increasing path in (thus if and only if there exists an increasing path in ), and

implies that .
Proof. First note that it is sufficient to prove the second item for length one paths (as longer paths are just concatenations of length one paths). We proceed by induction on , the number of couples such that , such that there exists an increasing path, and such that there is no increasing path. In the initial case, , as for every vertex such that there exists an increasing path, we are done and .
If , consider such couple with the property that and are consecutive in (by taking as the lowest element in such that there is no increasing path). Note that as otherwise, extending an increasing path with the edge one obtains an increasing path. By Lemma 10, the representation obtained by permuting and is such that , such that the orders between the endpoints of any edge are preserved (and thus the increasing paths are preserved), such that for every , such that for any two vertices and without increasing path nor path, if and only if , and has only couples .
We can thus apply the induction hypothesis to and we obtain that there exists a representation such that , such that the increasing paths are the same as in (and thus as in ), such that for every , such that if and only if there exists an increasing path in , and such that implies that , which implies (as there is no increasing path nor path).
Lemma 12
Let be a representation on . For any face and any vertex of , there exists an increasing path for some order , where is the maximal vertex of in order .
Proof. We proceed by induction on , the number of orders such that . If then , a contradiction. So the lemma holds by vacuity. If , consider an order such that . By Lemma 11 (applied to in ) either contains an increasing path, and we are done, or there exists a representation such that , such that the increasing paths are the same as in , such that for every , and such that . In this case, we apply the induction hypothesis on . Indeed, and has only orders such that . Note that as every pair in corresponds to an edge, the maximal vertices of in and are the same for every . The induction thus provides us an increasing path and this path is also increasing in .
Lemma 13
Let be a representation on . For any vertex set , there exists a vertex which does not dominate , and such that for every there exists an increasing path for some vertex .
Proof. We proceed by induction on the number of elements which do not dominate . If then , a contradiction. So the lemma holds by vacuity. If , consider any such element which does not dominate . By Lemma 11 either there are increasing paths for every and we are done, or there exists a representation such that , such that the increasing paths are the same as in , such that for some order , such that implies that , and such that for every .
In this case, we apply the induction hypothesis on . Indeed, as any element not dominating in does not dominate in , and as is not dominating in , we have at most elements that do not dominate in . By induction hypothesis there exists a vertex such that for every , has an increasing path for some vertex , and this path is also increasing in .
3 TDDelaunay complexes
TDDelaunay graphs have been introduced by Chew in [7]. Here we generalize this definition to higher dimensions. We recall that a positive homothety of is an affine transformation of defined by where and . The coordinates of a point will be denoted .
For any integer , let be the
dimensional hyperplane of
defined by . Given , we define a regular simplex of by setting . A regular simplex is said to be positive if . For we call the canonical regular simplex. In this context, a point set is in general position if for any two vertices , for every . The interior of a regular simplex is defined byProposition 14
The positive regular simplices of are the subsets of positively homothetic to .
Proof. Let be a subset of positively homothetic to . Let us show that there exists such that and . There exists a positive homothety of of ratio and center such that . We define by . We show that . If it is clear that and that . We can therefore suppose that .
Let . Then for every . Then . Furthermore . Then and thus .
Let . Then for every . Because , we define such that , that is such that . As , then . Furthermore for every , . Thus . So, .
We conclude that . Furthermore if then is positive. Indeed . We deduce that every subset of which is positively homothetic to is a positive regular simplex.
Let be a positive regular simplex of with such that . We look for and such that . Suppose that such an and an exist. Then for every . Then . Thus and . It is easy to check that this gives the desired and and that they are well defined even if or . We conclude that is positevly homothetic to in .
Let us now define TDDelaunay simplicial complexes by extending the notion of TDDelaunay graph defined by Chew and Drysdale [5].
Definition 15
Given a set of points of () in general position, let the TDDelaunay complex of , denoted , be the simplicial complex with vertex set defined as follows. A subset is a face of if and only if there exists a positive regular simplex such that and such that no point of is in the interior of .
Let , we define by . Remark that because for every , .
Lemma 16
For any set , if and only if does not contain any point of in its interior.
Proof. Suppose that . Then there exists a positive regular simplex such that and such that no point of is in the interior of . As contains , for every and every , . Thus for every and . Therefore does not contain any point of in its interior otherwise would contain some.
Suppose that does not contain any point of in its interior. Let . By definition of , for every . So and contains . Let . If is in , then is not in the interior of . So there exists such that . But there exists (different from ) such that . This contradicts the fact that the points of are in general position. Therefore and .
Proposition 17
For any point set in general position in (), is an abstract simplicial complex.
Proof. Consider any , and any . Then for every . So . does not contain any point of in its interior otherwise would contain some. Thus because of the previous lemma, and we conclude that is an abstract simplicial complex.
Consider a point set of in general position. We define the orders on as if and only if , in other words if and only if . First, note that as the points are in general position these orders are well defined. Note also that the values for and form a solution of a linear system of inequalities (with inequalities of the form ). In the following we connect TDDelaunay complexes to representations through systems of inequalities. To do so we consider these orders as a representation denoted . If this representation is closely related to .
Theorem 18
For any point set of in general position, we have that . Thus, any TDDelaunay complex of has DushnikMiller dimension at most .
Proof. Consider a set of points of in general position and let be the representation on such that if and only if .
Let us first prove that , by showing that for any we have that . By definition there exists such that contains exactly the points , and they lie on its border. For every , we denote by the maximum among the elements of with respect to . Towards a contradiction we suppose that . Thus there exists a vertex of such that does not dominate in any order of . Thus for every . Therefore for every . But because thus . Hence contradicting the definition of .
Let us now prove that , by showing that for any we have that . Consider any nonempty face (the case of the empty face is trivial) and suppose towards a contradiction that . According to Lemma 16, there exists such that . Thus for every . For every , we define as the maximum among the elements of with respect to . Thus , and for every , which contradicts the fact that .
The reciprocal statement holds for and 3, as any representation is such that defines a TDDelaunay complex of . This naturally raised the following conjecture [17] (See also in [11] as an open problem).
Conjecture 19 ([17])
For every representation , the abstract simplicial complex is a TDDelaunay complex of .
Mary [17] proved the conjecture when already admits some particular embedding. In the following we show that actually this conjecture does not hold, already for . To do so, in the following we characterize which representations are such that is a TDDelaunay complex. Actually we are first going to characterize which representations correspond to some point set of such that , and then we are going to show that for those we have . By definition of , for any different and for any order we have that , if and only if
(1) 
Furthermore as we consider a point set of we have that , which gives:
(2) 
In the following we consider a representation , and we define the system of inequalities obtained by taking Inequality (1) for every but only for the pairs , and by replacing the coordinates by the right hand of Equation (2).
Definition 20 (TDDelaunay system)
Let be a representation on a vertex set , and consider the edge set of defined by . We define the matrix of where the coefficients, of are indexed by an edge , a vertex , and two indices and .
The TDDelaunay system of the representation is the following linear system of inequalities :
for some vector
.Example 21
We consider the following representation on :
The complex is given by the facet and contains edges: and . The matrix of the TDDelaunay system of is:
The system where is equivalent to the following linear system, where denotes .
Note that setting , , and to , , and respectively, the last three equations imply that , and .
Theorem 22
For any abstract simplicial complex with vertex set , is a TDDelaunay complex of if and only if there exists a representation on such that and such that the corresponding TDDelaunay system has a solution.
Proof. () This follows from Theorem 18 and from the fact that the coordinates of any point set form a solution to the TDDelaunay system defined by .
() Consider now a representation on a set such that the TDDelaunay system of has a solution and for any let us define a point by setting for , and . This implies that all these points belong to . It will be clear from the context when we refer to an element of or to the corresponding point of . As in the linear system the inequalities are strict, one can slightly perturb the position of the vertices in order to obtain points in general position that still fulfill the system constraints. Recall that by construction, for any edge of and any , if and only if . This implies that if has an increasing path then .
Let us first prove that . Consider a face and suppose towards a contradiction that . As , there exists such that contains exactly the points , and they lie on its border. As , then by Lemma 13, there exists which does not dominate , and such that for every there exists an increasing path for some vertex . Therefore . As does not dominate and as the points are in general position one of these inequalities is strict and we conclude that lies in the interior of , a contradiction.
Let us now prove that . Consider any nonempty face . For every , we denote the maximum of in the order and we define (and ) by setting . First note that for any vertex and any , as is an edge, we have that . We hence have that . As , for every there exists an such that . Therefore and as for every (because either or and then ), we have that is on the border of . According to Lemma 12, for every , there exists an increasing path in , for some order . Therefore and . Thus <
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