Planar order on vertex poset

TL;DR

This paper introduces the concept of planar order on vertex posets of processive planar graphs, establishing a natural correspondence with the planar order on their edge posets, and extends the concept to general finite posets.

Contribution

It proves the existence of a natural planar order on vertex posets derived from the edge poset's planar order in processive planar graphs.

Findings

Planar order on vertex posets is equivalent to conjugate order.

A natural planar order on vertex posets can be induced from edge posets.

The concept extends to any finite poset, not just graphs.

Abstract

A planar order is a special linear extension of the edge poset (partially ordered set) of a processive plane graph. The definition of a planar order makes sense for any finite poset and is equivalent to the one of a conjugate order. Here it was proved that there is a planar order on the vertex poset of a processive planar graph naturally induced from the planar order of its edge poset.

Equations

$$i^{-}(v)=minI(v),$$

$$i^{+}(v)=maxI(v),$$

$$o^{-}(v)=minO(v),$$

$$o^{+}(v)=maxO(v).$$

Full text

Keywords: edge poset, vertex poset, planar order

1 Introduction

The notion of a processive plane graph, a special case of Joyal and Street’s progressive plane graph [2], was introduce in [3] as a graphical tool for tensor calculus in semi-groupal categories. In [3], we gave a totally combinatorial characterization of an equivalence class of processive graphs in terms of the notions of a POP-graph which is a processive graph (a special kind of acyclic directed graph) equipped with a planar order (a special linear order of the edges).

However, it turns out that the notion of a planar order can be defined for a general finite poset (partially ordered set) and essentially equivalent to the one of a conjugate order [1], which is an important notion in the study of planar posets. So this raises an interesting question: for a processive graph, are there some relations between planar orders on its edges and planar orders on its vertices? In this paper, we will give a positive answer to this question by showing that any planar order of edges of a processive graph naturally induces a planar order of vertices.

2 Processive plane graph

Definition 1**.**

A processive plane graph is an acyclic directed graph drawn in a plane box with the properties that: $(1)$ all edges monotonically decrease in the vertical direction; $(2)$ all sources and sinks are of degree one; and $(3)$ all sources and sinks are placed on the horizontal boundaries of the plane box.

Figure $1$ shows an example.

Figure $1$. A processive plane graph

Processive plane graphs can also be defined in terms of processive graphs [3] and their boxed planar drawings.

Definition 2**.**

A processive graph is an acyclic directed graph with all its sinks and sources of degree one.

A planar drawing of processive graph $G$ is called boxed [2] if $G$ is drawn in a plane box with all sinks of $G$ on one horizontal boundary of the plane box and all sources of $G$ on the other horizontal boundary of the plane box. A planar drawing of an acyclic directed graph is called upward if all edges increases monotonically in the vertical direction (or other fixed direction). Thus a processive plane graph is exactly a boxed and upward planar drawing of a processive graph.

Definition 3**.**

Two processive plane graphs are equivalent if they are connected by a planar isotopy such that each intermediate planar drawing is boxed (not necessarily upward).

In [3], equivalence classes of processive plane graphs are mainly used to construct a free strict tensor category on a semi-tensor scheme.

3 Planar order and POP-graph

In [3], we gave a combinatorial characterization of an equivalence classes of a processive plane graph in terms of a planar order on its underlying processive graph. In this paper, we define planar order for any poset.

Definition 4**.**

A planar order on a poset $(X,\rightarrow)$ is a linear order $\prec$ on $X$, such that

$(P_{1})$* for any $x_{1},x_{2}\in X$, $x_{1}\rightarrow x_{2}$ implies $x_{1}\prec x_{2}$;*

$(P_{2})$* for any $x_{1},x_{2},x_{3}\in X$, $x_{1}\prec x_{2}\prec x_{3}$ and $x_{1}\rightarrow x_{3}$ imply that either $x_{1}\rightarrow x_{2}$ or $x_{2}\rightarrow x_{3}$.*

$(P_{1})$ says that $\prec$ is a linear extension of $\rightarrow$.

Recall that two partial orders on a set are conjugate if each pair of elements are comparable by exactly one of them. It is easy to see that $(P_{2})$ is equivalent to the condition that if $e_{1}\prec e_{2}\prec e_{3}$, then $e_{1}\not\rightarrow e_{2}$ and $e_{2}\not\rightarrow e_{3}$ imply that $e_{1}\not\rightarrow e_{3}$. Thus $(P_{2})$ enables us to define a transitive binary relation: $e_{1}<e_{2}$ if and only if $e_{1}\prec e_{2}$ and $e_{1}\not\rightarrow e_{2}$; moreover, if $(P_{1})$ is satisfied, then the linearity of $\prec$ implies that $<$ is a conjugate order of $\rightarrow$. So the planar order $\prec$ is a reformulation of the conjugate order of $\rightarrow$.

In a directed graph, we denote $e_{1}\rightarrow e_{2}$ if there is a directed path starting from edge $e_{1}$ and ending with edge $e_{2}$. Similarly, $v_{1}\rightarrow v_{2}$ denotes that there is a directed path starting from vertex $v_{1}$ and ending with vertex $v_{2}$. For any acyclic directed graph, its edge set and vertex set are posets with the relation $e_{1}\rightarrow e_{2}$ and $v_{1}\rightarrow v_{2}$. We call them edge poset and vertex poset of the acyclic directed graph, respectively.

The following is a key notion in [3].

Definition 5**.**

A planarly ordered processive graph or POP-graph , is a processive graph $G$ together with a planar order $\prec$ on its edge poset $(E(G),\rightarrow)$.

We simply denote a POP-graph as $(G,\prec)$; see Fig $2$ for an example.

6$$5$$9$$12$$15$$8$$19$$1$$2$$3$$4$$10$$11$$17$$7$$13$$14$$16$$18

Figure $2$. A POP-graph

A basic result is the following.

Theorem 6** ([3]).**

There is a bijection between POP-graphs and equivalence classes of processive plane graphs.

The POP-graph in Fig $2$ corresponds to the processive plane graph in Fig $1$.

4 Planar order on vertices

In this section, we will prove our main result. Before that we need some preliminaries.

From now on, we fix be a POP-graph $(G,\prec)$. For a vertex $v$ of $(G,\prec)$, the set $I(v)$ of incoming edges and the set $O(v)$ of outgoing edges are linearly ordered by $\prec$. We introduce some notations when $I(v)$ or $O(v)$ are not empty:

[TABLE]

[TABLE]

[TABLE]

[TABLE]

The following lemma is a result first proved in [4].

Lemma 7**.**

Let $v$ be a vertex of $(G,\prec)$. If the degree of $v$ is not one, then $o^{-}(v)=i^{+}(v)+1$ under the linear order $\prec$.

Proof.

Notice that $G$ is a processive graph, then $deg(v)\neq 1$ implies that $I(v)\neq\emptyset$ and $O(v)\neq\emptyset$. Thus both $i^{+}(v)$ and $o^{-}(v)$ exist. Now we prove $o^{-}(v)=i^{+}(v)+1$ by contradiction. Suppose there exists an edge $e$, such that $i^{+}(v)\prec e\prec o^{-}(v)$. Since $i^{+}(v)\rightarrow o^{-}(v)$, then by $(P_{2})$ we have $i^{+}(v)\rightarrow e$ or $e\rightarrow o^{-}(v)$. If $i^{+}(v)\rightarrow e$, then there must exists an edge $e^{\prime}\in O(v)-\{o^{-}(v)\}$, such that $e^{\prime}\rightarrow e$ or $e^{\prime}=e$. Thus $e^{\prime}\preceq e$, which contradicts $e\prec o^{-}(v)$. Otherwise, $e\rightarrow o^{-}(v)$, then there must exist an edge $e^{\prime\prime}\in I(v)-\{i^{+}(v)\}$ such that $e\rightarrow e^{\prime\prime}$ or $e^{\prime\prime}=e$. Then $e\preceq e^{\prime\prime}$, which contradicts $i^{+}(v)\prec e$. ∎

Lemma 7 shows that for any vertex $v$, $\overline{E(v)}=\overline{I(v)}\sqcup\overline{O(v)}$, where $E(v)$ is the set of incident edges of $v$ and $\overline{X}$ denotes the interval of subset $X$ in a poset. Due to Lemma 7, we can define a linear order $\prec_{V}$ on the vertex set $V(G)$. For any two different vertices $v_{1},v_{2}$ of $G$, $v_{1}\prec_{V}v_{2}$ if and only if one of the following conditions is satisfied:

$(1)$ $I^{+}(v_{1})\prec I^{+}(v_{2})$, $(2)$ $I^{+}(v_{1})\prec O^{-}(v_{2})$, $(3)$ $O^{-}(v_{1})\preceq I^{+}(v_{2})$, $(4)$ $O^{-}(v_{1})\prec O^{-}(v_{2})$.

We write $v_{1}\preceq_{V}v_{2}$ if $v_{1}=v_{2}$ or $v_{1}\prec_{V}v_{2}$. The following Theorem is our main result.

Theorem 8**.**

For any POP-graph $(G,\prec)$, $\preceq_{V}$ defines a planar order on the vertex poset $(V(G),\rightarrow)$.

Proof.

$(1)$ $\prec_{V}$ satisfies $(P_{1})$. If $v_{1}\rightarrow v_{2}$, then there exist $e_{i}\in E(G)$ $(1\leq i\leq n)$ such that $v_{1}=s(e_{1})$, $v_{2}=t(e_{n})$ and $t(e_{i})=s(e_{i+1})$ for $(1\leq i\leq n-1)$, which implies that $o^{-}(v)\preceq e_{1}\preceq e_{n}\preceq i^{+}(v_{2})$. Thus $o^{-}(v_{1})\preceq i^{+}(v_{2})$, then by definition of $\prec_{V}$, we have $v_{1}\prec_{V}v_{2}$.

$(2)$ $\prec_{V}$ satisfies $(P_{2})$. Suppose $v_{1}\prec_{V}v_{2}\prec_{V}v_{3}$ and $v_{1}\rightarrow v_{3}$, then $o^{-}(v_{1})$ and $i^{+}(v_{3})$ exist and $o^{-}(v_{1})\prec i^{+}(v_{3})$. We have four cases:

Case 1: $v_{1}$ is a source and $v_{3}$ is a sink. In this case, notice that $G$ is processive, then by Definition $2$, $\{o^{-}(v_{1})\}=O(v_{1})$ and $\{i^{+}(v_{3})\}=I(v_{3})$. So $v_{1}\rightarrow v_{3}$ implies that $o^{-}(v_{1})\rightarrow i^{+}(v_{3})$. Let $e=i^{+}(v_{2})$ or $o^{-}(v_{2})$, then $v_{1}\prec_{V}v_{2}\prec_{V}v_{3}$ implies that $o^{-}(v_{1})\prec e\prec i^{+}(v_{3})$ or $o^{-}(v_{1})=e$ or $e=i^{+}(v_{3})$. In the first case, by $(P_{2})$, we have $o^{-}(v_{1})\rightarrow e$ or $e\rightarrow i^{+}(v_{3})$, which implies that $v_{1}\rightarrow v_{2}$ or $v_{2}\rightarrow v_{3}$. In the second case, we have $v_{1}\rightarrow v_{2}$, and in the third case, we have $v_{2}\rightarrow v_{3}$.

Case 2: $v_{1}$ is not a source and $v_{3}$ is a sink. In this case, $i^{+}(v_{1})$ exists and by Definition $2$, $\{i^{+}(v_{3})\}=I(v_{3})$. So $v_{1}\rightarrow v_{3}$ implies that $i^{+}(v_{1})\rightarrow i^{+}(v_{3})$. Let $e=i^{+}(v_{2})$ or $o^{-}(v_{2})$, then $v_{1}\prec_{V}v_{2}\prec_{V}v_{3}$ implies that $i^{+}(v_{1})\prec e\prec i^{+}(v_{3})$ or $e=i^{+}(v_{3})$. In the first case, by $(P_{2})$, we have $i^{+}(v_{1})\rightarrow e$ or $e\rightarrow i^{+}(v_{3})$, which implies that $v_{1}\rightarrow v_{2}$ or $v_{2}\rightarrow v_{3}$. In the second case, we have $v_{2}\rightarrow v_{3}$.

Case 3: $v_{1}$ is a source and $v_{3}$ is not a sink. This case is similar to case $(2)$.

Case 4: $v_{1}$ is not a source and $v_{3}$ is not a sink. In this case, both $i^{+}(v_{1})$ and $o^{-}(v_{2})$ exist and $v_{1}\rightarrow v_{3}$ implies that $i^{+}(v_{1})\rightarrow o^{-}(v_{3})$. Let $e=i^{+}(v_{2})$ or $o^{-}(v_{2})$, then $v_{1}\prec_{V}v_{2}\prec_{V}v_{3}$ implies that $i^{+}(v_{1})\prec e\prec i^{+}(v_{3})$. By $(P_{2})$, we have $i^{+}(v_{1})\rightarrow e$ or $e\rightarrow o^{-}(v_{3})$, which implies that $v_{1}\rightarrow v_{2}$ or $v_{2}\rightarrow v_{3}$. ∎

Figure $3$ shows the planar order on the vertex poset of the POP-graph in Fig $2$.

2$$3$$4$$1$$7$$12$$13$$15$$8$$9$$16$$18$$5$$6$$11$$10$$14$$17$$19$$20

Figure $3$. Induced planar order on vertices.

Theorem 8 shows that for any processive graph, each conjugate order of its edge poset induces a conjugate order of its vertex poset. However, in general, the converse is not true. Therefore, together with Theorem 2.1, Theorem 3.1 demonstrates that edge poset is more effective tool than vertex poset in the study of upward planarity. It is worth to mention that Fraysseix and Mendez, in a different but essentially equivalent context, also showed a similar judgement (in their final remark of [1]). In our subsequent work, we will show that for a transitive reduced processive graph (directed covering graph of a poset), a planar order on its vertex set can naturally induce a planar order on its edge set, which is essentially related the work in [1].

Acknowlegement

This work is supported by "the Fundamental Research Funds for the Central Universities".