By Kondratev A.S.
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4 Let = n − 1 n . 18) ⎪ ⎪ ⎩ 1 1 1 − i2 vi − i vi+1 if j = i + 1. Let M be the natural permutation Sn -module with basis e1 the irreducible submodule N = i ai ei ∈ M i ai = 0 . Set vj = 1 e1 + · · · + ej−1 − j − 1 ej j j −1 en . 18). Let ∈ n and ∈ V n − k . Set / = HomSn−k V resSn−k V It is clear from the branching rule that V / = 0 if and only if the Young diagram is contained in the Young diagram , in which case we denote the complement by / . A set of nodes of this form will be called a skew shape.
2 Irreducible representations of Sn are defined over are self-dual. 3 (Young’s orthogonal form) Let ∈ n . There exists a GZ-basis wT of V such that the action of an arbitrary simple transposition si ∈ Sn is given by si wT = wT + 1 − −1 where = res Ti+1 − res Ti (note that when is zero, so this term should be omitted). 1, and set w T = vT / vT vT Let S = si T . We may assume that si is an admissible transposition and S > T . 1(ii) imply vS vS = 1 − 2 vT vT , whence wS = vS / vT vT 1 − 2 . 13). 4 Let = n − 1 n .
So now the induction hypothesis on n gives that all such coefficients cL are zero, as required. 2 If L is an irreducible module in n -mod, then L L. Proof Since xi = xi , leaves characters invariant. Hence it leaves irreducibles invariant, since they are determined up to isomorphism by their character according to the theorem. 3 Let M ∈ Suppose: m -mod and N ∈ N indn+m M; (i) indm+n mn M nm N m+n ind (ii) M N appears in resm+n mn M mn Then indm+n mn M n -mod be irreducible modules. N with multiplicity one.