Generalized Foulkes Conjecture and Tableaux Construction

Transcription

1 Generalized Foulkes Conjecture and Tableaux Construction Thesis by Rebecca Vessenes In Partial Fulfillment of the Requirements for the egree of octor of Philosophy California Institute of Technology Pasadena, California 2004 (Submitted May 27, 2004)

2 194 Appendix A Association between Tableaux Spaces and Irreducibles Let n = ab and H = S b S a. Recall that W λ,a = {T T a λ-tableau filled with 1 to a, each b times}. We will explicitly show why the multiplicity of χ λ in 1 Sn H dimension of C{q T T W λ,a }. View H as a subgroup of S n, where H acts on equals the 1, 2,... b b + 1,..., 2b... (a 1)b + 1,..., ab by S b on each block and by S a permuting the blocks. The elements of H are the form (π 1,... π a, σ) with π i S b, σ S a. Now S a S n acts on W λ,a with S a acting on the numbers 1 to a and S n acting on the positions (corresponding to labelling across the rows). Let = {(σ 1, (π 1,... π a, σ)) π i S b, σ S a }. So S a H S a S n. Let T be the λ-tableau filled across the rows with b 1 s, then b 2 s, etc. Then S a S n acting on T gives W λ,a and fixes T. Specifically, Stab Sa S n (T ) =. Hence as S a S n modules, W λ,a 1 S a S n Stab(T ) = 1S a S n. Proposition A.0.2. W λ,a µ a ϕ S a (µ) (ϕ H (µ)) S n where ϕ Sa (µ) is the irreducible of S a indexed by µ and ϕ H (µ) is the irreducible of H/(S b... S b ) S a indexed by µ. Proof. Since 1 (S b... S b ), it is in the kernel of 1. As S b... S b H, it

3 195 is in the kernel of 1 S a H. So we can view 1 S a H as an S a H/(S b... S b ) S a S a module. Let = {(σ 1, σ) σ S a } be the image of in S a H/(S b... S b ). Hence 1 Sa H 1 Sa Sa as S a H/(S b... S b ) modules. Thus we can write 1 Sa Sa = aµ,ν φ µ φ ν for µ, ν a and some a µ,ν, where φ is the corresponding irreducible of S a. By Frobenius reciprocity a µ,ν = (φ µ φ ν, 1 Sa Sa ) = (φ µ φ ν, 1 ). Now φ µ φ ν = φ µ φ ν. So (φ µ φ ν, 1 ) = 1 σ S a φ µ (σ)φ ν (σ) = Sa (φ µ, φ ν ) Sa. Using 1 if µ = ν row orthogonality and S a =, we have a µ,ν =. 0 otherwise So 1 S a S a = µ a φ µ φ µ. If we lift back to the original module 1 S a H have 1 Sa H = µ φ S a (µ) φ H (µ) where φ Sa (µ) is the irreducible of S a indexed by µ and φ H (µ) is the irreducible of H/(S b S b ) S a indexed by µ. Since 1 S a S n = (1 S a H ) Sa Sn we get we 1 S a S n = ( µ φ Sa (µ) φ H (µ)) S a S n = µ φ Sa (µ) (φ H (µ)) S n. By this proposition we have W λ,a µ a ϕ S a (µ) (ϕ H (µ)) Sn as S a S n modules. Consider the submodule on which S a is trivial, that is, µ = (a). This corresponds to 1 Sa (1 H ) S n. If 1 Sn H = ν n m νχ ν, this module corresponds to ν n 1 S a m ν χ ν. Now e λ = σ R T τ C T ɛ(τ)στ is an idempotent of S n on λ-tableau T. So the action of e λ on 1 m ν χ ν is the same as the action of q λ = π S a πe λ on W λ,a. Then q λ W λ,a m λ (e λ S λ ) as S n modules, as e λ S ν = 0 for λ ν. Now S λ is is a cyclic S n -module generated by e λ (T ). (Correspondingly, the semi-standard tableaux which span S λ are equivalent under the action of S n.) Therefore dim(e λ S λ ) = 1 and dim(q λ W λ ) = m λ. Hence {q T T W λ,a }, spans a module of dimension m λ, the multiplicity of χ λ in 1 Sn S b S a. This proof is due to Wales, [22].

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