By Mary Carson

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**Sample text**

We have the following relations: γ˜x,y,z = γ˜y,z,x (a) ∑ γ˜x−1 ,y,w n˜w = δxy (b) for all x, y, z ∈ W , for all x, y ∈ W , w∈W n˜w = n˜ w−1 and γ˜x,y,z = γ˜y−1 ,x−1 ,z−1 (c) for all w, x, y, z ∈ W . Proof. (a) Just note that the defining formula for γ˜x,y,z is symmetrical under cyclic permutations of x, y, z. (b) Using the defining formulae for γ˜x,y,z and n˜ w , the left-hand side evaluates to ∑ ∑ w∈W ∑ λ ∈Λ s,t,u∈M(λ ) = ∑ tu us fλ−1 cst x−1 ,λ cy,λ cw,λ ∑ ∑ λ ,μ ∈Λ s,t,u∈M(λ ) v∈M(μ ) ∑ ∑ μ ∈Λ v∈M(μ ) tu f λ−1 f μ−1 cst x−1 ,λ cy,λ f μ−1 cvv w−1 , μ vv ∑ cus w,λ cw−1 ,μ .

Then we perform a slight transformation as follows. We set σˆ εj (Tw ) := P−1 σ εj (Tw )P (w ∈ W ), where P := 1 − 12 0 1 . Then Ωˆ j σˆ εj (Tw−1 ) = σˆ εj (Tw )tr Ωˆ j for all w ∈ W , where Ωˆ j := Ptr Ω j P. Furthermore, 24 1 Generic Iwahori–Hecke Algebras Ωˆ j ≡ 20 0 12 Ωˆ j ≡ 0 2(2 + ζ j + ζ − j ) 1 j −j 0 2 (2 − ζ − ζ ) if L(s1 ) > L(s2 ) = 0, mod m if L(s1 ) = L(s2 ) = 0. 5 applies again and so σˆ εj is balanced. But then σ εj must also be balanced, since the transforming matrix P has all its entries in K.

From the explicit knowledge of cλ one can deduce explicit formulae for the invariants aλ and f λ . If L(s) = 0 for all s ∈ S, then cλ = |W |/dλ . Hence, aλ = 0 and fλ = |W |/dλ for all λ ∈ Λ in this case. Now assume that L(s) > 0 for at least some s ∈ S. For W of exceptional type H3 , H4 , E6 , E7 , E8 (where we are automatically in the equal-parameter case), see the tables in [220, Chap. 4] and in the Appendices C and E in [132]. 2 (p. 3 Lusztig’s a-Invariants 17 diagram, we can assume without loss generality that L(s1 ) = L(s2 ) For the types I2 (m), An−1 , Bn and Dn , see the examples below.