Density and Fractal Property of the Class of Oriented Trees

03/23/2019 ∙ by Jan Hubička, et al. ∙ Universitat Politècnica de Catalunya Charles University in Prague 0

We show the density theorem for the class of finite oriented trees ordered by the homomorphism order. We also show that every interval of oriented trees, in addition to be dense, is in fact universal. We end by considering the fractal property in the class of all finite digraphs.



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1. Introduction

In this note we consider finite directed graphs (or digraphs) and countable partial orders. A homomorphism between two digraphs is an arc preserving mapping from to . If such homomorphism exists we write . The relation , called the homomorphism order, defines a quasiorder on the class of all digraphs which, by considering equivalence classes, becomes a partial order.

In the past two decades the richness of the homomorphism order of graphs and digraphs has been extensively studied [3]. In 1982, Welzl showed that undirected graphs, with one exception, are dense [7]. Later in 1996, Nešetřil and Zhu characterized the gaps and showed a density theorem for the class of finite oriented paths [6]. We contribute to this research by showing a density theorem for the class of oriented trees. We say that an oriented tree is proper if its core is not a path.

Theorem 1.1.

Let and be two finite oriented trees satisfying . If is a proper tree, then there exists a tree such that .

This result was claimed by Miroslav Treml around 2005, but never published. Our proof of Theorem 1.1 is new and simple, and leads to further consequences. In particular, we can show the following strengthening. Let us say that a partial order is universal if it contains every countable partial order as a suborder.

Theorem 1.2.

Let and be two finite oriented trees satisfying . If is a proper tree, then the interval is universal.

Recently, it has been shown that every interval in the homomorphism order of finite undirected graphs is either universal or a gap [2]. As consequence of Theorem 1.2, this property, called fractal property, seems to be present in other classes of digraphs. In fact, we have shown the following result related to the class of finite digraphs.

Theorem 1.3.

Let and be two finite digraphs satisfying , where the core of is connected and contains a cycle. Then the interval is universal.

The proof of Theorem 1.3 will appear in the full version of this note.

2. Preliminaries

In this note we shall use the notation used in Hell and Nešetřil’s book [3].

A digraph

is an ordered pair of sets

where is a set of elements called vertices and is a binary irreflexive relation on . The elements , denoted , of are called arcs.

A path is a digraph consisting in a sequence of different vertices together with a sequence of different arcs such that is an arc joining and for each . A cycle its defined analogously but with . A tree is a connected digraph containing no cycles. The height of a tree is the maximum difference between forward and backward arcs of a subpath in it.

A homomorphism from a digraph to a digraph is a mapping such that implies . It is denoted . If there exists a homomorphism from to we write , or equivalently, . We shall write for and . The interval consists in all digraphs such that . A gap is an interval in which there is no digraph such that .

The relation is clearly a quasiorder which becomes a partial order by choosing a representative for each equivalence class, in our case the so called core. A core of a digraph is its minimal homomorphism equivalent subgraph.

Given two partial orders and , an embedding from to is a mapping such that for every , if and only if . If such a mapping exists we say that can be embedded into .

Finally, a partial order is universal if every countable partial order can be embedded into it.

3. Density Theorem

In order to prove Theorem 1.1 we shall construct a tree from a given proper tree which will satisfy for every tree .

Given a tree , a vertex and a set of vertices , the plank from to , denoted , is the subgraph induced by the vertices of every path which starts with and contains some vertex .

Let be the core of a proper tree. Then there exists a vertex such that is adjacent to at least three different vertices, name them . Without loss of generality we shall assume that and are arcs. Let be the set of vertices, different from and , which are adjacent to . Note that is not empty since . Let , and . Observe that . See Figure 2.

Figure 1. Tree .
Figure 2. Tree .
Figure 3. Tree . Observe the enumeration of the vertices and planks of each tree .

Now, let be the tree from Figure 2, where and are copies of the plank , is a copy of , and and are copies of .

Finally, let be a tree consisting in consecutive trees whose planks are identified with the planks of the following trees. See Figure 3. We shall refer to the vertices for as labelled vertices.

Lemma 3.1.

Let and be finite oriented trees such that is a proper tree and . If there exists a homomorphism , then every labelled vertex of is mapped to a different vertex of .


Assume that is a core and consider a homomorphism . Observe that two consecutive labelled vertices can not be mapped via to the same vertex since it would imply that contains a loop. Now, observe that if any pair of labelled vertices of distance two are mapped to the same vertex, it will induce a homomorphism . This follows from the construction of . See Figure 3. Finally, if two labelled vertices at distance greater or equal to three are mapped to the same vertex, it would imply that contains a cycle, which is a contradiction since is a tree. We conclude that every labelled vertex has to be mapped to a different vertex of . ∎

A digraph is rigid if it is a core and the only automorphism is the identity. We shall use the following fact.

Fact 3.2.

The core of a tree is rigid.

Proof of Theorem 1.1 (sketch).

Assume that is a core. Let and consider the tree . It is clear that . It can also be checked that (here we might use Fact 3.2). To see that observe that by Lemma 3.1 every labelled vertex in has to be mapped to a different vertex in , but the number of labelled vertices in is greater than . Thus, .

We end by joining with by a proper and long enough zig-zag. The method is similar to the one use in the proof of the density theorem for paths [6]. ∎

4. Fractal property for proper trees

Proof of Theorem 1.2 (sketch).

Let and consider the tree . We know by Theorem 1.1 that .

Let be the core of . By Lemma 3.1 every labelled vertex in has to be mapped to a different vertex in . Since , it follows that has at least labelled vertex. Let and be the initial and ending labelled vertex of respectively. Let be the tree obtained from by adding two new vertices and and joining to and to by a proper zig-zag of length 5 or 6 so and have the same level, as shown in Figure 4. Finally let be the tree obtained by joining with by a proper and long enough zig-zag.

Figure 4. This is an example of how might look. The vertices and might be different from the ones in the figure but they must be labelled vertices of .

Now, we shall construct an embedding from the homomorphism order of the class of oriented paths, which we know is a universal partial order [1], into the interval .

Given an oriented path , let be the tree obtained by replacing each arc in by a copy of identifying with and with . Observe that . It is clear that any homomorphism induces a homomorphism by identifying arcs with copies of . To see the opposite, observe that since is rigid by Fact 3.2, every copy of in must be map via the identity to some copy of in . It follows that adjacent copies of in must be mapped to adjacent copies of in . Hence, each homomorphism induces a homomorphism . ∎

5. Fractal property for finite digraphs

We say that a class of digraphs has the fractal property if every interval in the homomorphism order is either universal or a gap. The fractal property was introduced by Nešetřil [4] and it has been shown recently that the class of finite undirected graphs (or symmetric digraphs) has the fractal property [2]. In this note, we have shown that the class of proper trees has also the fractal property (as consequence of Theorem 1.2). However, the class of finite digraphs, and even the class of oriented trees, is more complicated.

Nešetřil and Tardif characterised all gaps in the homomorphism order of finite digraphs [5]. It was shown that for every tree there exists a digraph such that is a gap, and that all gaps have this form. Theorem 1.1 contributes to this result by implying that if is a gap and is a proper tree, then must contain a cycle.

The characterisation of universal intervals in the homomorphism order of finite digraphs seems to be complicated. Related to this issue, we have stated Theorem 1.3. Its proof combines some techniques already used [3, 2] with some extra arguments, and will appear in the full version of this note. This result together with Theorem 1.2 imply that the class of finite digraphs whose cores are not paths has the fractal property. However, intervals of the form , where the core of is a path, remain to be studied and characterised.


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