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TauCeti.RepresentationTheory.Quiver.Reflection.Forest

Any two orientations of a forest are related by reflections at sinks #

Reflecting a quiver at a sink reverses the arrows meeting that vertex and keeps its underlying multigraph (Quiver.nonempty_reflectList_sum_hom_equiv). This file proves the converse for a forest: if a quiver q on a finite vertex type has at most one arrow between any two vertices and an acyclic underlying graph, then every quiver q' with the same underlying multigraph is reached from q by reflecting along a sink-admissible list of vertices (Quiver.exists_isSinkAdmissible_reflectList_equiv).

The arrows are turned around one at a time. Removing an arrow u ⟶ w from a forest disconnects w from u; let B be the set of vertices still connected to w. The arrow u ⟶ w is the only one joining B to its complement, and it points into B, so no arrow leaves B. Reflecting at the vertices of B, listed so that no arrow runs from an earlier entry to a later one, is then sink-admissible, and it reverses exactly the arrows joining B to its complement: the single arrow u ⟶ w. The orientation of a forest is acyclic (TauCeti.Quiver.isAcyclic_of_isAcyclic_underlyingGraph), which provides such a listing.

Main results #

References #

theorem Quiver.exists_isSinkAdmissible_reflectList_equiv {V : Type u} [Finite V] (q q' : Quiver V) (hsub : ∀ (a b : V), Subsingleton ((a ⟶ b) ⊕ (b ⟶ a))) (hG : (TauCeti.Quiver.underlyingGraph V).IsAcyclic) (h : ∀ (a b : V), Nonempty ((a ⟶ b) ⊕ (b ⟶ a) ≃ (a ⟶ b) ⊕ (b ⟶ a))) :
∃ (l : List V), TauCeti.Quiver.IsSinkAdmissible q l ∧ ∀ (a b : V), Nonempty ((a ⟶ b) ≃ (a ⟶ b))

Any two orientations of a finite forest are related by reflections at sinks. Let q be a quiver on a finite vertex type with at most one arrow between any two vertices, counted in both directions, whose underlying graph is acyclic. If a second quiver q' has the same arrows joining any two vertices, in either direction, then reflecting q along some sink-admissible list of vertices produces a quiver whose arrows are those of q'.