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Bruhat covers are right multiplication by positive-root reflections

Statement

Let x,y∈W with y⊲x a cover. Then there is a unique positive root β∈Φ+ with x=ysβ, and ℓ(x)=ℓ(y)+1. Conversely, if β∈Φ+ and ℓ(ysβ)=ℓ(y)+1, then ysβ⊳y is a cover. In the notation of the Bruhat graph the label β of the arrow x→y is characterised by sβ=y−1x, and it is also the unique positive root γ with y=xsγ and ℓ(xsγ)=ℓ(x)−1. The proof uses the standard sign criterion ℓ(wsγ)<ℓ(w)  ⟺  wγ<0 together with the reflection-chain description of Bruhat order.

Facts & Assumptions

Given: The finite reduced crystallographic root system Φ with positive system Φ+, the Weyl group W with simple reflections, and the Bruhat order.

[F1]

u≤v in Bruhat order is equivalent to the existence of a saturated reflection chain u=w0,w1,…,wk=v with wj+1=tjwj for root reflections tj and ℓ(wj+1)=ℓ(wj)+1; every such chain has exactly ℓ(v)−ℓ(u) steps, so u≤v with ℓ(v)=ℓ(u)+1 means v=tu for a root reflection t (Bruhat order on a finite Weyl group).

[F2]

For a positive root γ∈Φ+ and its reflection sγ: ℓ(sγw)<ℓ(w) if and only if w−1γ<0, and ℓ(wsγ)<ℓ(w) exactly when wγ<0; only multiplication by a simple reflection is guaranteed to change length by one. Also ℓ(w−1)=ℓ(w) (Finite Weyl strong exchange and deletion).

[F3]

Root reflections are the maps sβ for roots β; sβ=s−β, and sβ=sγ forces γ=±β because the only scalar multiples of a root in Φ are ± itself; W permutes the root set (Root reflections and the Weyl group action, Finite Weyl root system, lattice and chamber conventions).

Proof

1.1F1F2F3algebra

Let y⊲x. By [F1] with a one-step chain, x=ty for a root reflection t; write t=sγ with γ a root and replace γ by −γ if necessary so that γ∈Φ+. Since x=sγy and ℓ(x)=ℓ(y)+1>ℓ(y), the criterion in [F2] applied to w=y forbids y−1γ<0; hence β:=y−1γ∈Φ+. Conjugation gives x=y (y−1sγy)=y sy−1γ=ysβ with β=y−1γ∈Φ+.

2.1F3step 1.1

Suppose x=ysβ=ysβ′ with β,β′∈Φ+. Then sβ=sβ′, so β′=±β by [F3], and positivity forces β′=β. Thus the positive root in step 1.1 is unique, and multiplying x=ysβ on the left by y−1 gives y−1x=sβ, so the label is determined by the group elements.

2.2F1F3step 1.1

Conversely let β∈Φ+ and suppose ℓ(ysβ)=ℓ(y)+1. Put t:=ysβy−1=syβ by conjugation, so ysβ=ty is a one-step saturated reflection chain; by [F1], y≤ysβ. If z satisfied y<z<ysβ, then by [F1] any saturated chain from y to ysβ through z would have more than one step, so ℓ(ysβ)−ℓ(y)≥2, contradicting the hypothesis. Hence ysβ covers y.

3.1step 1.1step 2.1step 2.2∎

Finally, if γ∈Φ+ satisfies y=xsγ and ℓ(xsγ)=ℓ(x)−1, then x=ysγ by multiplying on the right by sγ, and ℓ(ysγ)=ℓ(x)=ℓ(y)+1; step 1.1 applied to the cover y⊲x (whose existence is the hypothesis y=xsγ) gives γ=β. Step 2.1 supplied the uniqueness of β from the pair (x,y) alone, so the two characterisations of the label coincide.

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