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The Kazhdan–Lusztig inversion formula

Facts & Assumptions

Given: n≥1 and the finite standard and Kazhdan–Lusztig bases of Hv(n).

[F2]

The chain-defined matrix Q′ is the two-sided inverse of P, and the classical sign matrix S gives Q=SQ′S (Inverse Kazhdan–Lusztig polynomials).

[F3]

Bruhat intervals in Sn are finite (Basic properties of the Bruhat order on Sn).

Statement

Let P=(px,w), Q′=(qx,w′) and R=(rx,y) be the triangular matrices of Existence and uniqueness of the Kazhdan–Lusztig basis, Inverse Kazhdan–Lusztig polynomials and The R-coefficient recursion, support, degree bounds and inversion. Then (a) P=RPˉ, Pˉ=RˉP, and RRˉ=RˉR=1; (b) Q′P=PQ′=1, and the inversion formulas qx,w′‾=∑x≤z≤wqx,z′rz,w,qx,w‾=∑x≤z≤wqx,zrz,w‾ hold, i.e. Qˉ′=Q′R and Qˉ=QRˉ; (c) for the dual basis Dx(H‾w)=δx,w from Inverse Kazhdan–Lusztig polynomials, Dx(Hw)=qx,w′. The inversion is proved from the bar-duality relations alone (no finite case check).

Proof

technique · compare the coefficient matrices and use the chain inverse defining $Q'$
1.1F1algebra

Bar-duality matrices. Entrywise bar applied to P=RPˉ gives Pˉ=RˉP, because bar is an involution and is multiplicative on matrices over the commutative coefficient ring. The R theorem gives both RRˉ=I and RˉR=I. Thus (a) holds.

2.1F1F2F3step 1.1algebra

The inverse matrix identity. By [F2], Q′=P−1. From P=RPˉ, inversion gives Q′=Pˉ−1R−1=Qˉ′Rˉ, since Pˉ−1=P−1‾=Qˉ′ and R−1=Rˉ. Applying entrywise bar yields Qˉ′=Q′R. Its (x,w) entry is qx,w′‾=∑zqx,z′rz,w; triangular support restricts this finite sum to x≤z≤w.

3.1F1F2F3step 2.1algebra

The signed inverse formula. Let S be diagonal with Sx,x=sgn(x), so S2=I. From [F2], Q=SQ′S; from the R bar-symmetry, Rˉ=SRS. Therefore Qˉ=SQˉ′S=SQ′RS=(SQ′S)(SRS)=QRˉ, whose (x,w) entry is qx,w‾=∑zqx,zrz,w‾. Triangular support again restricts to x≤z≤w.

4.1F1F2F3step 1.1step 2.1algebra∎

Dual-basis interpretation. Since (H‾w) is a basis of the finite free module Hv(n), its coordinate functionals Dx form the dual basis and satisfy Dx(H‾w)=δx,w. Put dx,w:=Dx(Hw). Evaluating on H‾w=∑zpz,wHz gives DP=I; because P is invertible, D=P−1=Q′, so Dx(Hw)=qx,w′. Conversely, if the dual evaluations are qx,w′, the identity Q′P=I gives Dx(H‾w)=δx,w. This proves (c). All sums are finite by [F3], and no choice principle is used.

Remarks

The matrix inversion uses the locally proved basis/bar-duality clauses and the finite chain inverse with its sign convention. Coefficientwise positivity is not required.

Depends on

Used by

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Sources