A Geometric Approach to Acyclic Orientations

The set of acyclic orientations of a connected graph with a given sink has a natural poset structure. We give a geometric proof of a result of Jim Propp: this poset is the disjoint union of distributive lattices. Let G be a connected graph on the vertex set [n] = {0} ∪ [n], where [n] denotes the set {1, . . . , n}. Let P denote the collection of acyclic orientations of G, and let P0 denote the collection of acyclic orientations of G with 0 as a sink. If Ω is an orientation in P with the vertex i as a source, we can obtain a new orientation Ω with i as a sink by firing the vertex i, reorienting all the edges adjacent to i towards i. The orientations Ω and Ω agree away from i. A firing sequence from Ω to Ω in P consists of a sequence Ω = Ω1, . . . ,Ωm+1 = Ω ′ of orientations and a function F : [m] −→ [n] such that for each i ∈ [m], the orientation Ωi+1 is obtained from Ωi by firing the vertex F (i). We will abuse language by calling F itself a firing sequence. We make P into a preorder by writing Ω ≤ Ω if and only if there is a firing sequence from Ω to Ω. From the definition it is clear that P is reflexive and transitive. While P is only a preorder, P0 is a poset. By finiteness, antisymmetry can be verified by showing that firing sequences in P0 cannot be arbitrarily long. This is a consequence of the fact that neighbors of the distinguished sink 0 cannot fire. The proof depends on the following lemma. Lemma 1. Let F : [m] −→ [n] be a firing sequence for the graph G. If i and j are adjacent vertices in G, then |F(i)| ≤ |F(j)| + 1. Proof. A vertex can fire only if it is a source. Firing the vertex i reverses the orientation of its edge to the vertex j. Hence the vertex i cannot fire again until the orientation is again reversed, which can only happen by firing j. As a corollary, firing sequences have bounded length, implying that P0 is a poset. Corollary 2. The preorder P0 of acyclic orientations with a distinguished sink is a poset. Proof. Let F : [m] −→ [n] be a firing sequence. By iterating the lemma, |F−1(i)| ≤ d(0, i) − 1, so