Iwahori-Hecke algebras and their representation theory: by Ivan Cherednik, Yavor Markov, Roger Howe, George Lusztig,

By Ivan Cherednik, Yavor Markov, Roger Howe, George Lusztig, Dan Barbasch, M. Welleda Baldoni

Uncomplicated difficulties of illustration concept are to categorise irreducible representations and decompose representations occuring evidently in another context. Algebras of Iwahori-Hecke sort are one of many instruments and have been, most likely, first thought of within the context of illustration idea of finite teams of Lie kind. This quantity contains notes of the classes on Iwahori-Hecke algebras and their illustration conception, given through the CIME summer time tuition which came about in 1999 in Martina Franca, Italy.

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Extra resources for Iwahori-Hecke algebras and their representation theory: lectures given at the C.I.M.E. summer school held in Martina Franca, Italy, June 28-July 6, 1999

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K Then 5 becomes n−2k+1 1 12 2 yk q q Tk q − q−1 q q n−2k+1 2 Combining powers of q, dividing by q n−2k 2 yk = q n−2k−1 2 1 yk+1 q 2 Tk . , and rearranging produces 1 1 Tk yk − yk+1 Tk = (q 2 − q − 2 )yk . Using the action sk (yk ) = yk+1 , this can be rewritten in the form of the Bernstein-Zelevinsky relation 1 1 Tk yk − sk (yk )Tk = (q 2 − q − 2 ) sk (yk ) − yk . sk (yk )yk−1 − 1 (6) Note also that Tk yj = yj Tk for j = k, k + 1 since then sk commutes with aj and aj−1 as observed above. In addition, the yj generate an abelian subalgebra and the Tk satisfy the relations Ti T j = Tj T i Tk Tk+1 Tk = Tk+1 Tk Tk+1 if |i − j| > 1 for 1 ≤ k ≤ n − 1.

Notice then multiply throughout by fa−1 k k−1 that fak−1 is indeed invertible. ) This yields −1 1 q fsk (fak fak−1 ) − 1− 1 q (fak fa−1 ) = fa−1 fak+1 fsk = fak+1 fa−1 fsk . k−1 k k In equation 5, replace fw by q 1 Tk = q − 2 fsk , yk = q − − (w) 2 n−2k+1 2 (5) fw and rename by setting fak fa−1 , k−1 and yk+1 = q − n−2k−1 2 fak+1 fa−1 . k Then 5 becomes n−2k+1 1 12 2 yk q q Tk q − q−1 q q n−2k+1 2 Combining powers of q, dividing by q n−2k 2 yk = q n−2k−1 2 1 yk+1 q 2 Tk . , and rearranging produces 1 1 Tk yk − yk+1 Tk = (q 2 − q − 2 )yk .

For 1 ≤ a ≤ n (where dim V = 2n), let Λa be the lattice spanned by {ej , 1 ≤ j ≤ n} ∪ {fj , a < j ≤ n} ∪ {πfj , 1 ≤ j ≤ a}. Let Λn+a be the lattice spanned by {ej , 1 ≤ j ≤ n − a} ∪ {πej , n − a < j ≤ n} ∪ {πfj , 1 ≤ j ≤ n}. Then set Λb+2nc = π c Λb for 0 ≤ b < 2n, and c ∈ Z. The reader can check that the Λm form a complete lattice flag, which is self-dual. Precisely, Λ∗m = Λ−m . We will investigate the structure of self-dual lattice flags in V . We will find in particular that any complete self-dual lattice flag is symplectically equivalent to the example just given.

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