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Given an acyclic directed graph $G$, the set $E(G)$ of edges of $G$ equipped with the reachable order $\to$ is called the edge poset of $G$, where for two edges $e_1\to e_2$ means that there is a directed path starting from $e_1$ ending with $e_2$.

My question is that how can we characterize those edge posets of acyclic directed graphs, or under exactly what conditions a poset can be represented as the edge poset of an acyclic directed graph?

A closely related result is stated in the paper "Planarity and Edge Poset Dimension", at page 733, where the authors said that:

*Posets which are edge partial orders of graphs have been fully characterized (see , for example , [13]) . They are called N-free partial orders as they do not contain any N-shaped configuration . They are also characterized by the property that any maximal chain intersects any maximal antichain . We have the following property :

PROPERTY 3 . Each N - free partial order is the edge - partial order of an unique e - bipolarly oriented graph.*

This result means that the edge poset of an e-bipolarly oriented graph (or equivalently, an st-graph /PERT-graph) can be characterized by the property of N-freeness.

But I do not agree with the statement that posets which are edge partial orders of graphs have been fully characterized, which is what my question concerns.

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  • $\begingroup$ Isn't this the transitive closure of the DAG? en.wikipedia.org/wiki/Transitive_closure $\endgroup$ Commented May 15, 2023 at 8:07
  • $\begingroup$ @MatthieuLatapy No. The elements of the "edge poset" are the edges, not the vertices. For instance, the $4$-element poset whose Hasse diagram looks like the letter N can not be represented as an edge poset. $\endgroup$
    – bof
    Commented May 15, 2023 at 8:14
  • $\begingroup$ @Matthieu Latapy Yes, the reachable order of vertices is exactly the transitive closure of the adjacency relation of a DGA. But my question is about the reachable order of edges, not of vertices. $\endgroup$
    – xuexing lu
    Commented May 15, 2023 at 8:17

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The result that each N - free partial order is the edge - partial order of an unique e - bipolarly oriented graph, implies that a poset is an edge poset if and only if it is N-free.

Proof:$(\Rightarrow)$ We prove that edge posets are N-free by contradiction. Suppose $P$ is the edge-poset of acyclic directed graph $G$ and is not N-free, then it contains a covering-subposet $N$ with four elements $\{a,b,c,d\}$ such that $a$ covers $b$, $c$ covers $b$, $c$ covers $d$, $a||c$, $a||d$ and $b||d$. Evidently, the edges conrresponding to $a,b,c,d$ and their incident vertices form a sub acyclic directed graph of $G$ (an edge-induced subgraph), whose edge-poset is $N$. This contradicts with the fact that $N$ can not be an edge-poset.

$(\Leftarrow)$ This direction is a direct consequence of the fact that any N-free poset is the edge poset of an unique PERT-graph.

The conrrespondence between N-free posets and acyclic directed graphs are one-to-many. In fact, we can freely combine and split the sources and sinks of an acyclic directed graph, respectively, without changing its edge-poset. It is natural to introduce the following notion.

Two acyclic directed graphs are EP-equivalent if they have the same edge-poset.

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