LemmaStatement: AI-generatedProof: AI-generated

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AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
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Both-sided divisors of the longest positive lift

Statement

Let $(W, S)$ be finite, let $M$ be the presented positive Artin monoid, and let $w_{0}$ be the longest Coxeter element. Put $Δ = w_{0}$ using the monoid reduced-word lift. A left divisor of $Δ$ means $x \in M$ for which $x y = Δ$ for some $y \in M$ ; a right divisor means $y x = Δ$ for some $y$ . The left divisors and the right divisors of $Δ$ coincide, and both sets are exactly ${w : w \in W}$ . These elements generate $M$ and form a finite set.

Statement references

Positive degree and atoms of the presented Artin monoid

Reduced Coxeter lifts in the presented positive monoid

The longest element as the opposition of the chamber, and longest elements of finite parabolics

Facts & Assumptions

[F1]

The longest element satisfies $ℓ (w_{0} w) = ℓ (w w_{0}) = ℓ (w_{0}) - ℓ (w)$ and $w_{0}^{2} = 1$ . Word reversal gives $ℓ (w^{-} 1) = ℓ (w)$ . (The longest element as the opposition of the chamber, and longest elements of finite parabolics, part (1))

[F2]

Monoid lifts are well defined, have degree equal to Coxeter length and multiply along length-additive products. (Reduced Coxeter lifts in the presented positive monoid)

[F3]

The monoid degree is additive and equals the length of every representing word. (Positive degree and atoms of the presented Artin monoid)

Proof

Given: The objects and hypotheses in the statement.

Goal: Establish all assertions of the statement.

1.1F1F2

For any $w \in W$ , [F1] gives $ℓ (w^{- 1} w_{0}) = ℓ (w_{0}) - ℓ (w)$ and $ℓ (w_{0} w^{- 1}) = ℓ (w_{0}) - ℓ (w)$ . Applying [F2] to $w (w^{- 1} w_{0}) = w_{0}$ and $(w_{0} w^{- 1}) w = w_{0}$ proves that $w$ is both a left and a right divisor of $Δ$ .

2.1F2F3step 1.1∎

Conversely, suppose $x y = Δ$ . Choose words $X, Y$ representing $x, y$ . By [F3] their concatenation has length $d (x) + d (y) = d (Δ) = ℓ (w_{0})$ , and its image in $W$ is $w_{0}$ . Thus it is a reduced word for $w_{0}$ . Each of its initial and final subwords is reduced: a shorter expression for either would shorten the whole word. Consequently $X$ and $Y$ are reduced expressions for their images, and [F2] gives $x = q (x)$ and $y = q (y)$ . The same argument applies to $y x = Δ$ . Together with step 1.1 this identifies both divisor sets. They are finite since $W$ is finite, and they contain each generator $a_{s} = s$ , hence generate $M$ .

Depends on

Used by · 1

Independent spherical Artin Garside input

Dependency tree · 2 levels

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