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The Bernstein cross-relation quotients are Laurent polynomials

Statement

Let R=(X,Φ,Y,Φ∨,Δ) be the reduced based root datum and let C[Y] and the reflection action be as in The formal Bernstein presentation for a reduced based root datum. For every α∈Δ and y∈Y,

θy−θsα(y)1−θ−α∨∈C[Y]if α∉2X,

and

θy−θsα(y)1−θ−2α∨∈C[Y]if α∈2X.

Facts & Assumptions

Given: A reduced based root datum R=(X,Φ,Y,Φ∨,Δ), a simple root α∈Δ, and an element y∈Y.

[F1]

The root datum has an integral perfect pairing, and its simple reflection acts by sα(y)=y−⟨α,y⟩α∨ with ⟨α,y⟩∈Z (Abstract root data and their Weyl groups).

[F2]

The group algebra C[Y] has basis {θz:z∈Y} and multiplication θzθz′=θz+z′ (The formal Bernstein presentation for a reduced based root datum).

Proof

Given: The data above. Put n=⟨α,y⟩∈Z and u=θ−α∨.

1.1F1F2givenalgebra

By [F1], sα(y)=y−nα∨, so by [F2] the numerator is θy(1−un). If n>0, then (1−un)/(1−u)=1+u+⋯+un−1; if n=0, the quotient is zero; and if n=−k<0, then (1−un)/(1−u)=−u−k(1+u+⋯+uk−1). Each expression is a finite Laurent polynomial, proving the first assertion.

2.1F1F2step 1.1algebra∎

If α∈2X, write α=2β with β∈X. Then n=⟨2β,y⟩=2⟨β,y⟩=2m for an integer m. With u=θ−α∨, the numerator is θy(1−u2m). For m>0, (1−u2m)/(1−u2)=1+u2+⋯+u2m−2; for m=0 it is zero; and for m<0 the same finite negative-power identity from step 1.1, applied to u2, is a Laurent polynomial. Since 1−u2=1−θ−2α∨, the second assertion follows.

Depends on

Used by

Cited to discharge well-definedness by The formal Bernstein presentation for a reduced based root datum.

Dependency tree · two levels

8 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources