Vertex-Facet Assignments For Polytopes

12/20/2018 ∙ by Thomas Jahn, et al. ∙ TU Chemnitz 0

For polytopes in R^d with at least as many facets as vertices, we prove that vertices can be mapped injectively to non-incident facets when 1≤ d≤ 5 and give counterexamples for d≥ 7.

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1. Background and notation

Polytopes are present in many fields of mathematics such as poset theory, discrete mathematics, linear optimization, and convex geometry. Unsurprisingly, problems involving polytopes can be addressed by a variety of methods, and different aspects of such problems can be emphasized. A motivation for the contents of the present paper is given by a metrical notion called reducedness introduced for general convex sets in [5]. While the existence of reduced polytopes is clear in the Euclidean plane, the only known examples of reduced polytopes in higher-dimensional Euclidean spaces are those given in [3]. This existence problem for a polytope with a certain metrical property has a combinatorial feature at its core: The authors of [1, Theorem 4] show that, in particular, for any reduced polytope, each vertex of the polytope is (in some sense) opposite to a facet to which it is not incident, and this assignment is injective. Thus, are there combinatorial types of polytopes in for which their vertices cannot be mapped injectively to non-incident facets and therefore cannot have any reduced realizations? In the following, we proof that the answer is negative for and affirmative for .

Let us recall some standard notation. For this, we refer to the books [4, 7]. A polytope in is the convex hull of finitely many points , i.e.

By its dimension, we mean the dimension of its affine hull, or if is empty. A face of is a subset with the property, that for all and , the condition already implies . The faces of include the empty set and itself. Each face is also a polytope and therefore has a dimension. The family of -dimensional faces of is finite and shall be denoted by . Any two faces of which are not disjoint are said to be incident to each other. Throughout this paper, we use the standing assumption that the polytope is of dimension , and that this dimension is (we exclude the empty and singleton polytope). In this case, its -dimensional faces are called vertices, edges and facets when , and , respectively. The set consists of singletons and we will not distinguish between the point and the singleton . In particular, we will write . A polytope is spanned by its vertices: . A face of a face of and the intersection of faces of is again a face of . In particular, any face of is spanned by some vertices in . These will be called the vertices of the face. Our central problem now reads as follows:

Problem 1.1.

Given a polytope with , is there an injective mapping for which and are non-incident for all ?

Obviously, the assumption cannot be dropped as it is necessary for the existence of injective mappings without any extra condition. Polytope duality, which interchanges the roles of -dimensional faces and -dimensional faces, carries over any proof or counterexample to or from the following equivalent problem.

Problem 1.2.

Given a polytope with , is there an injective mapping for which and are non-incident for all ?

The injective maps mentioned in Problems 1.1 and 1.2 shall be called vertex-facet assignments and facet-vertex assignments, respectively.

2. Graph-theoretic formulation

Problems involving injective maps between finite sets can be stated conveniently in the language of graph theory, using matchings of bipartite graphs. We refer the reader to [2, Chapter 2]. In order to avoid confusion with the language of polytopes, we refer to the nodes of a graph, instead of its vertices. A bipartite graph is then a graph for which there is a decomposition of the set of nodes into non-empty disjoint sets called partition classes, such that edges (the elements of ) do not join nodes of the same partition class. For each node , denote by the neighborhood of , i.e. the set of to adjacent nodes in . Further, set . A matching in a bipartite graph is a -regular subgraph of . The matching of is said to be -covering (resp. -covering) if every node (resp. ) is a node in . In this sense, -covering matchings of correspond to injective single-valued selections of the set-valued mapping given by , that is, -covering matchings of are injective mappings which map nodes to adjacent ones.

For our purposes, a bipartite graph can be assigned to any polytope as follows.

Definition 2.1.

The vertex-facet graph of the polytope is the bipartite graph whose partition classes are (disjoint copies of) the vertices and the facets of the polytope. A vertex and a facet are adjacent in if and only of they are non-incident in .

With this terminology in place, the vertex-facet assignments (resp. facet-vertex assignments) are just -covering (resp. -covering) matchings in the vertex-facet graph.

Hall’s marriage theorem [2, Theorem 2.1.2] gives a necessary and sufficient condition for the existence of -covering matchings of bipartite graphs .

Theorem 2.2.

The bipartite graph possesses a -covering matching if and only if for all .

The condition for all appearing in Theorem 2.2 is called Hall condition. While one direction of the theorem is almost trivial, the other one gives the power to construct the desired matchings (provided they exist).

3. Existence in low dimensions

The answers to Problems 1.1 and 1.2 are trivially affirmative for . Trial and error for polytopes in with low facet or vertex count does not yield any counterexamples. Using Hall’s marriage theorem (Theorem 2.2) for the vertex-facet graph (see Definition 2.1), we are able to show that there are no counterexamples up to dimension . For the proof, we will take the dual point of view, i.e. we address Problem 1.2 which requires finding facet-vertex assignments for polytopes . This is motivated by the following geometric interpretation of the non-neighborhood of a set of nodes in the vertex-facet graph .

Proposition 3.1.

Let be the vertex-facet graph of the polytope . For , the set is the set of vertices of .

Proof.

Note that a vertex is in if and only if there is at least one facet to which it is non-incident (since then they are adjacent in ). Hence, a vertex lies in the complement if and only if it incident to all the facets , i.e. it lies in the intersection . ∎

In the sequel, we denote by and the vertex count and the facet count of the polytope under consideration. Also, we denote by the vertex-facet graph of . As a corollary of Theorem 2.2 and Proposition 3.1, we obtain the following statement.

Corollary 3.2.

A polytope possesses a facet-vertex assignment if and only if for all and facets , we have

(1)

In order to construct a counterexample to Problem 1.2, it is required to find a polytope and facets such that . The possible existence of counterexamples is constrained by the affine dimension of .

Theorem 3.3.

Let , be a polytope with and . If , then condition 1 holds.

Proof.

We deal with each case for separately.

Case : This means that is a vertex of . Assume that condition 1 is violated, i.e. . This implies , and we conclude , i.e. . Thus the intersection uses all facets of . But no vertex of is incident to all facets of , a contradiction.

Case : This means that is an edge of . Assume that condition 1 is violated, i.e. . This implies , and we conclude , i.e. . The affine hull of the edge

is a straight line which is contained in the hyperplanes obtained as the affine hulls of the facets

. Two additional hyperplanes (or facets of ) are required to “cut out” the edge from its affine hull. However, since , there is at most one facet of left which is not one of . This is a contradiction.

Case : This means that is a polygon with, say, vertices and edges. Assume that condition 1 is violated, i.e. . This implies , and we conclude , i.e. . Thus the intersection uses all but at most facets of . However, in order to “cut out” the edges of the polygon, we need more facets not already used in the intersection, a contradiction.

Case : This means that is a -face of . As such, it is incident to exactly two facets of , see [7, Theorem 2.7]. Hence . Assume that condition 1 is violated, i.e. . Thus the two facets and must intersect in all vertices, or all but one. Since the intersection is -dimensional, the convex hull of the union of and the (possibly) remaining vertex of is at most -dimensional. This contradicts the assumption of being -dimensional.

Case : This means that the intersection consists of a single facet, i.e. . Assume that condition 1 is violated, i.e. . Hence, the facet contains all vertices, which contradicts the assumption of being -dimensional. ∎

An immediate consequence of Theorem 3.3 is the existence of facet-vertex assignments for polytopes up to dimension .

Corollary 3.4.

Every polytope of dimension possesses a facet-vertex assignment.

Therefore, the smallest possible counterexample to Problem 1.2 might be of dimension , where some facets intersect in a face of dimension .

4. Counterexamples in high dimensions

In this section, we construct for any polytopes which do not possess vertex-facet assignments. This leaves open only the case of dimension . Our counterexamples are based on a standard construction for polytopes. Given two polytopes , their free join

is defined as the convex hull of these polytopes after embedding them into skew affine subspaces of

. If the vertices of the two polytopes are given as , then the free join can be defined as

Most importantly, if the polytopes have vertices and facets, then their free join has vertices and facets. This is quite obvious for the vertices. For the facets note the following: each facet of is spanned either by and a facet of , or by and a facet of . For details regarding the free join and its properties see [6, Corollary 2].

Theorem 4.1.

For , let be a polytope with vertices and facets. Assume that and , and that . Then does not possess any vertex-facet assignment.

Proof.

Note that , i.e.  has more facets than vertices, and this is precisely the minimal requirement for the existence of vertex-facet assignments. Let be the facets of which contain (there are many because one for each facet of ). Then

The claim now follows from Theorem 2.2 and Corollary 3.2. ∎

A more direct reasoning (without using the detour over the matching formulation) might go like this: In a vertex-facet assignment, each vertex originally from must be assigned to a unique non-incident facet. The only facets not incident to are the ones spanned by and a (non-incident) facet of . However, since , there are not enough facets of to provide such non-incident facets in for all vertices of .

Note that the condition is not used in the proof but follows from and anyway. Furthermore, the conditions and cannot be satisfied for polytopes in dimension . This is the reason why this construction only yields counterexamples for Problem 1.1 in dimensions .

Example 4.2.

We list some concrete counterexamples for .

  1. [label=(),leftmargin=*,align=left,noitemsep]

  2. For odd dimensions

    , let be any polytope with more vertices than facets, and let be the dual of . The free join then is a self-dual -dimensional polytope, see [6, Corollary 2]. By Theorem 4.1, the polytope does not possess vertex-facet assignments, and by self-duality, it does not possess facet-vertex assignments either.

  3. We can construct counterexamples for Problem 1.1 from any polytope with and . Take to be a bipyramid with vertices. Then and . The free join then satisfies the assumptions of Theorem 4.1, and thus is a polytope without vertex-facet assignments in dimension .

5. Open problems

Question 5.1.

Does every -dimensional polytope possess a vertex-facet assignment, or are there any counterexamples?

As noted in Section 3, any possible counterexample in six dimensions must have facets that intersect in a -dimensional face with more than vertices.
Acknowledgements. The second author gratefully acknowledges the support by the funding of the European Union and the Free State of Saxony (ESF).

References

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