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The dual trace basis for the type-A2 Hecke algebra

Example

Take the equal-parameter Hecke algebra of W=S3 over R=Z[v±1], with basis

1,T1,T2,T12,T21,Tw0,T12=T1T2,T21=T2T1,Tw0=T1T2T1=T2T1T2.

For the canonical trace τ(Tw)=δw,1, the inverse map on W is

1−1=1,s1−1=s1,s2−1=s2,(s1s2)−1=s2s1,(s2s1)−1=s1s2,w0−1=w0.

Thus the Gram matrix (τ(TxTy))x,y∈W in the displayed ordered basis is

(100000010000001000000010000100000001)

The trace-dual basis is (1,T1,T2,T21,T12,Tw0), so the dual vector to Tw is Tw−1. It is not generally the algebra inverse: if a=v−v−1, then T1−1=T1−a, whereas the trace-dual vector to T1 is T1.

Facts & Assumptions

Given: The equal-parameter type-A2 Hecke algebra over R=Z[v±1], its standard basis, and the coefficient-at-identity trace.

[F1]

For finite W, the canonical trace sends ∑wrwTw to r1. (The canonical Hecke trace and symmetrizing form)

[F2]

For all x,y∈W, τ(TxTy)=δx−1,y, and the trace-dual basis is {Tw−1:w∈W}. (The canonical Hecke trace is symmetric and perfect)

[F3]

R=Z[v±1] is the Laurent polynomial ring and Ti2=(v−v−1)Ti+1. (Universal parameters, the generic Coxeter Hecke algebra and generator conjugacy)

[F4]

The standard elements Tw form an R-basis, with multiplication TxTy=Txy for additive lengths. (The standard basis of the generic Hecke algebra and base change, Reduced-word independence of T_w and the length-multiplication rules)

Verification

Given: The data in Facts & Assumptions.

1.1F4given

The reduced words for the six elements of S3 give exactly the listed standard basis. Reversing each reduced word gives the displayed inverse map; in particular w0=s1s2s1 is its own inverse.

2.1F1F2step 1.1

By [F1], τ is the coefficient-at-identity functional; applying [F2] to the ordered basis in step 1.1 gives τ(TxTy)=1 exactly when the column index is y=x−1, and 0 otherwise. The resulting matrix is the displayed permutation matrix.

3.1F2step 2.1algebra

The matrix in step 2.1 permutes the coordinate basis by x↦x−1, so it has an inverse equal to itself and determinant −1, a unit in R. Its dual vectors are therefore Tx−1; directly, τ(TxTy−1)=δx,y.

4.1F3step 3.1algebra∎

The quadratic relation in [F3] gives T12=aT1+1, so T1(T1−a)=1=(T1−a)T1 and T1−1=T1−a. The Laurent ring R has distinct basis monomials v and v−1, hence a=v−v−1≠0; thus T1−1≠T1, while step 3.1 shows T1 itself is trace-dual to T1. This distinguishes the two notions explicitly.

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