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Empty-subset wall crossings are simple reflections

PDF p. 11: Remark 3.16

Statement

Let (W,S) be a Coxeter system with simple generators si indexed by the vertices i of a Coxeter diagram Γ. For K⊆Γ0, write WK=⟨sk:k∈K⟩. An associate of J⊆Γ0 is a subset I for which WI is conjugate to WJ, as in Associates of a parabolic subset. The empty subset is its only associate.

For every a∈Γ0, the connected-support notation of Connected component notation [I,a] gives [∅,a]={a}, the group W[∅,a] is cyclic of order two, and the simple wall-crossing element of Simple wall-crossing elements and maps is va,∅=sa.

Proof

Given: the Coxeter system and vertex a in the Statement.

[F1] The generators satisfy si2=1, and in the standard geometric representation siαi=−αi for a nonzero basis vector αi. Word length is the least length of a word in the simple generators. Coxeter-system, roots, length and weak-order conventions

[F2] The standard parabolic WK is generated by the sk for k∈K; in particular W∅={1}. Standard parabolic subgroups and parabolic roots

[F3] The set [I,a] is the component containing a in the diagram induced on I∪{a}. Connected component notation [I,a]

[F4] For finite WL, wL is its maximum-length element; w∅=1. Longest parabolic elements and opposition permutation

[F5] A pair (x,I) lies in Cham(J) exactly when WJx=xWI and x has minimum word length in xWI. For such a label, if a∈Γ0∖I, D=[I,a] and WD is finite, the wall-crossing definition gives va,I=wD∖{a}wD. Algebraic chamber-label set Simple wall-crossing elements and maps

Proof technique: direct.

Goal: prove the associate, component, group-order and wall-crossing assertions.

1.1F1F2given

Associates. By [F1], si acts nontrivially on αi, so si≠1 for every vertex i. By [F2], a nonempty I has si∈WI for some i∈I, so WI is nontrivial. Since a conjugate of W∅={1} is again {1}, no nonempty I is an associate of ∅. The empty subset is an associate of itself by conjugation by 1.

1.2F1F2F3F4

Rank one. The induced diagram on {a} has only the vertex a, so [F3] gives D=[∅,a]={a}. By [F1] and [F2], WD={1,sa} and its two elements are distinct; hence it is cyclic of order two. In particular, WD is finite. The identity has length zero, while the nonidentity generator sa has length one. Thus [F4] gives wD=sa and wD∖{a}=w∅=1.

2.1F1F2F5givenstep 1.1step 1.2∎

Chamber label and wall crossing. Set J=I=∅ and x=1. By [F2], W∅1=1W∅={1}. The identity is the only element in this coset and has word length zero by [F1], so it has minimum length there. The chamber-label criterion in [F5] therefore gives (1,∅)∈Cham(∅). Also a∈Γ0∖∅ by the given choice of vertex. Step 1.2 establishes the remaining finiteness hypothesis, so [F5] applies and gives va,∅=w∅w{a}=1sa=sa. Together with steps 1.1 and 1.2 this proves the Statement.

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