By I. G. Macdonald

This new and lots more and plenty elevated variation of a well-received publication continues to be the one textual content to be had with regards to symmetric features and corridor polynomials. There are new sections in virtually each bankruptcy, and plenty of new examples were incorporated all through.

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Then for each integer r > 0 we have n-1 Mn+r= r- ( -1)"S(rIp)(X1.... ,Xn)M"-p-1 P-0 (If Mntr=EapMn-p-1, we have these equations for a0, . )) 16. Let A, µ be partitions of length n, and let PP(A, FA,) = det(PA;+µj+2n-i-j)lti,jtn' 50 SYMMETRIC FUNCTIONS I with the understanding that po = n. Then in A we have sAS, = P (A, µ) /P (0, 0). ) (Observe that 17. (a) Let p be an integer > 2 and let (o = e2"`1P. , wp-1) = ± 1 if A 0 (§1, Example 8(f)) and is zero otherwise. , WP-') = vP(A), where op(A) is the sign e(w) of the unique permutation w E SP such that A + SP w8P(mod.

X,,, it is therefore divisible by AS(x I a) in R[x1, ... , x,, ]. As in the text, we may assume that a1 > a2 > ... e. that a = A + 8 where A is a partition of length < n.

X. 6) For any a = (a1, ... , an) E Nn, let Aa = (x70, Ha = (ha;-n+j) (n x n matrices). Then Aa = Ha M. Proof Let n-1 e(k)tr = Fj (1 + xit). E(k)(t) _ r-O i,'k Then H(t)E(k)( -t) = (1-Xkt)-1, 42 SYMMETRIC FUNCTIONS I By picking out the coefficient of t" on either side, we obtain n F, ha;-n+j . ( -1) n-j (k) _ a, j-1 en-j -xk and hence HaM=A,,. 6): we obtain as = det(Aa) = det(H,,)det(M) for any a e Nn, and in particular det M = as, since det(H8) = 1. 8) W(SA) = SA. for all partitions A. t t It should be remarked that if R, R' are raising operators, RR'hA = hRR'A is not necessarily equal to R(R'hA).

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