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- The element κn is the class sum corresponding to the conjugacy class of the element s12 in Gr,p,n and thus κr,n is a central element of ℂGr,p,n. The constant by which κr,n acts on Gr,p,nλj follows from theorem???.
- The first statement follows from parts (a) and (b) and Theorems 3.6 and 3.22 as follows. By theorem 3.6, ℂAkn≅EndSnV⊗k if n≥2k. Thus, by Theorem 3.22, if n≥2k then Zk acts on the irreducible ℂAkn-module Akλn by the constant given in the statement. This means that Zk is a central element of ℂAkn for all n≥k. Thus, for n≥2k,dZk=Zkd for all diagrams d∈Ak. Since the coefficients in dZk (in terms of the basis of diagrams) are polynomials in n, it follows that dZk=Zkd for all n∈ℂ.
If n∈ℂ is such that ℂAkn is semisimple let χℂAknλ be the irreducible characters. Then Zk acts on Akλn by the statement, therefore it is a polynomial in n, determined by its values for n≥2k.
The proof of the second statement is completely analogous using ℂAk+12,Sn-1, and the second statement in part (b).
- Then 2κnvi1⊗…⊗vik=1r∑m=0r-1∑i,j=1,i≠jntimtj-msijvi1⊗…⊗timtj-msijvik=1r∑m=0r-1∑i,j=1,i≠jn1-Eii-Ejj+timEij+tj-mEjivi1⊗…⊗1-Eii-Ejj+timEij+tj-mEjivik. Expanding this sum, let cS,I=∏l∈Scδilil'-1#ll'⊆I+#ll'⊆Icξml∈Il'∈Ic-#l∈Il∈Ic∏l∈Iδili∏l∈Icδilj and
cS,I=∏l∈Scδilil'-1#ll'⊆I+#ll'⊆Icξml∈Il'∈Ic-#l∈Il∈Ic∏l∈Iδili∏l∈Icδilj so that 2κnvi1⊗…⊗vik is equal to 1r∑S⊆1…k∑i1',…,ik'∑m=0r-1∑i,j=1,i≠jn∑I⊆S∪S'cS,Ivi1'⊗…⊗vik'=1r∑S⊆1…k∑i1',…,ik'∑m=0r-1∑I⊆S∪S'∑i,j=1ncS,Ivi1'⊗…⊗vik'-∑i,j=1ncS,I'vi1'⊗…⊗vik'. Here Sc⊆1…k corresponds to the tensor products where 1 is acting, I⊆S∪S' corresponds to the tensor position that must equal i, and Ic corresponds to the tensor positions that must equal j.
When S=0 the set I is empty and the sum in (???) is equal to n2-nvi1⊗…⊗viksincecS,I=cS,I'=∏l∈1…kδilil'. Assume S≥1 and separate the sum according to the cardinality of I. Note that the sum for I is equal to the sum for Ic since the whole sum is symmetric in i and j. When I=S∪S', cS,I=cS,I'=∏l∈Scδilil'-1S∏<S∪S'δili and the sum in (???) is equal to 1r∑m=0r-1∑i1',…,ik'n-1∑i=1ncS,Ivi1'⊗…⊗vik'=n-1-1SbSvi1⊗…⊗vik. We get a similar contribution from the sum of the terms with I=∅.
For each of the remaining subsets I⊆S∪S' the sum in (???) contributes 0 when κI≠0modr and -1#ll'⊆I+#ll'⊆IcdI⊆Svi1⊗…⊗vik=-1S+κIdI⊆S-bSvi1⊗…⊗vik. when κI=0modr.
For the second statement, since 1-Eii-Ejj+EiiEjjvn=0,ifi=n or j=n,vn,ifotherwise,2κn-1vi1⊗…⊗vik⊗vn=1r∑m=0r-1∑i,j=1,i≠jn-1timtj-msijvi1⊗…⊗vik⊗vn=1r∑m=0r-1∑i,j=1,i≠jntimtj-msijvi1⊗…⊗timtj-msijvik⊗1-Eii-Ejj+EiiEjjvn=1r∑m=0r-1∑i≠jsijvi1⊗…⊗vik⊗vn+1r∑m=0r-1∑i,j=1,i≠jn1-Eii-Ejj+ξmEij+ξ-mEjivi1⊗……⊗1-Eii-Ejj+ξmEij+ξ-mEjivik⊗-Eii-Ejjvn=1r∑m=0r-1∑i,j=1ntimtj-msijvi1⊗…⊗timtj-msijvik⊗EiiEjjvn. The first sum is known to equal 2κnvi1⊗…⊗vik and is known by the computation proving the first statement, and the last is 0 since i≠j. The middle sum is treated exactly as in (???) except that now the sum is over S such that k+1∈S and I such that k+1k+1'⊆I or k+1k+1'⊆Ic.□
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