证明
Lemma 5 ([Induction from coordinate sectors]). Let a finite group G act…Proposition 4. If m≥1 , λ N_m , and ρ_mλ , then S^λ occurs in W_m .观点
Projection and the high positions. Use the ordinary tensor identification E_m^ ⊗B_m ⊗E_m^ ⊗B_m ≅(E_m⊗E_m)^ ⊗B_m . Put E_ lo =(e_1,…,e_s) and E_ hi =(e_ s+1 ,…,e_m) . On a pair-letter basis vector define p_s(e_a⊗e_b)= e_a⊗e_b,, a,b≤s or a,b>s,; 0,, otherwise, P_s=p_s^ ⊗B_m . The same local projection is used at every position, so P_s commutes with G . Call a retained position high when both letters are greater than s . In a retained word of w_m , let A be the set of high positions. Alternation in the first layer gives row counts r_i=(h+1-i,0) , and alternation in the second gives column counts d_j=(h+1-j,0) , for 1≤i,j≤m . For each 1≤k≤m , these counts satisfy ∑_ i=1 ^k r_i =# (i,j)∈B_h:i≤k =∑_ j=1 ^m(k,d_j). Column j can contribute at most (k,d_j) high positions to the first k rows. Equality of the sums in therefore forces equality in every column. For d_j>0 , take k=d_j : all its d_j high positions occupy the first d_j rows. Columns with d_j=0 have no high positions. Consequently A= (i,j)∈B_m:i≤h+1-j =B_h=D. For the zero-one matrix recording membership in A , this is the equality case of the row and column margin inequalities . The specific staircase margins in make every prefix bound an equality. Factorization with signs. In a row or column block of length r , its intersection with D is its initial segment of length a=(r-s,0) ; its intersection with F has length b=(r,s) . The retained assignments put the high letters s+1,…,r on the first segment and the low letters 1,…,b on the second. Let denote the unnormalized alternation of a displayed list, placed in the indicated segment, with (∅)=1 . The part of the block tensor with this prescribed high set is exactly (-1)^ ab \, (e_ s+1 ,…,e_r)⊗(e_1,…,e_b). Here the high list is empty when r≤s . Indeed, every high letter precedes every low letter in the position order, giving ab inversions; all other inversions are internal to the two independently permuted lists. Every such pair of permutations occurs once. In particular, when r<s the formula is just the original alternation on e_1,…,e_r . Rows and columns both have the length list m,m-1,…,1 . Thus their cross signs in multiply to ∏_ r=1 ^m(-1)^ 2(r-s,0)(r,s) =1. Identify E_h with E_ hi by e_a↦e_ s+a , and write w_h and W_h for the images of w_h and W_h in the two high-letter layers on D . The low portions are precisely the blocks of R_F,C_F . Regrouping the factors over the disjoint sets D,F introduces no signs, and hence z:=P_s(w_m)=w_h⊗u_ m,s ≠0. The nonvanishing follows also directly from the nonzero alternating tensors on each disjoint block. The two factors of H=S_D×S_F act independently on this tensor, while each acts diagonally on the two layers over its own position set. Therefore A_0:=[H]z , = (σw_h)⊗(τu_ m,s ): σ∈S_D, τ∈S_F ; , =W_h U_ m,s ≅W_h U_ m,s . The second equality follows by expanding tensors of linear combinations of the two separate orbit sets. The coordinate sectors. Let Ω be the set of subsets A⊂B_m of size |D| . Within the pair-letter space define T_A=(E_ hi ⊗E_ hi )^ ⊗A ⊗ (E_ lo ⊗E_ lo )^ ⊗(B_m A) . These are the coordinate spans of words high exactly on A . They have disjoint word bases, so their sum is direct, and gT_A=T_ gA . By, z∈T_D ; the stabilizer of D in G is exactly H . Apply Lemma 5 ([Induction from coordinate sectors]) and to obtain [G]z≅_H^G A_0 ≅_H^G(W_h U_ m,s ). Since P_s is equivariant, P_s(W_m)=[G]P_s(w_m)=[G]z . Its restriction to W_m , followed by the inverse map Ψ in and the identification of A_0 in , gives. The image of this ambient projection need not lie in the two original Specht factors; the composition is still a surjective G -equivariant linear map onto the displayed module. Semisimplicity over proves the last assertion.