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By Pannenberg M.

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It follows easily from this lemma that that n is at least spanned by all x w, ∈ n+ w ∈ Sn . 2 The x w ∈ n + w ∈ Sn form a basis for n. Proof We give two proofs of this important result. First proof. 7). 3). 2] we see that ˜ n has a basis given by all x˜ w words w ˜ in the s˜j which do not involve subword of the form s˜j2 . Hence, the subalgebra ˜ n of ˜ n generated by the x˜ i is isomorphic to n . Also let ˜n denote the subalgebra of ˜ n generated by the s˜j , so that ˜n is isomorphic to the algebra on generators s˜1 s˜n−1 subject to relations s˜j2 = 1 for each j.

Next, forget about A, and take care of n − 1 . Continuing this way we will get a chain of admissible transpositions which transform T to T . Finally, note that this chain yields a reduced word. 9 If and ≈ . ∈ W n and ∼ for some ∈ W n , then ∈W n Proof Let = T . 7, = S . As ∼ , the tableaux S and T have the same shape. 3(iii), it suffices to show that we can go from S to T by a chain of admissible transpositions. 8. The following is the main result of this chapter. 10 We have W n = W n Moreover, T ≈ S ⇔ the tableaux T and S have the same shape ⇔ T ∼ S .

1 Define a linear map w = n →d by d−1 w if w ∈ dS , 0 otherwise, Degenerate affine Hecke algebra 32 for each w ∈ Sn . Then: (i) is a homomorphism of -bimodules; (ii) ker contains no non-zero left ideals of n ; (iii) the map f n → Hom is an isomorphism of h→h d n -bimodules. n Proof (i) Follows easily using S d = dS . (ii) It is enough to show that n t = 0 for any t ∈ n . By multiplying t with an appropriate group element on the left, we may assume that t = y∈D yhy with each hy ∈ and hd = 0. Now t = hd = 0, as required.

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