<< . .

. 10
( : 11)



. . >>

n≥0 Λn (B)t for the generating function for elementary symmetric functions, which
are related by »(B; ’t)σ(B; t) = 1. Then, by [95, Def. 4.7 and Prop. 4.15], we have
σ((1 ’ q)B; 1) = »(B; ’q)σ(B; 1). Let X be the ordered alphabet · · · < q 2 < q < 1, so
that XA = A/(1 ’ q). According to [95, Theorem 4.17], it then follows that
σ((1 ’ q)A; 1) — σ(XA; 1) = σ((1 ’ q)XA; 1) = »(XA; ’q)σ(XA; 1)
= »(XA; ’q)σ(XA; q)σ(A; 1) = σ(A; 1),
since by de¬nition of X, σ(XA; 1) is equal to σ(XA; q)σ(A; 1) (see [95, Def. 6.1]).
Therefore, Sn ((1 ’ q)A) — Sn (XA) = Sn(A), as required.
Hence, we infer that Kn (q) is the inverse of Sn ((1 ’ q)A)/(q; q)n.
The eigenvalues of Sn ((1 ’ q)A) are given in [95, Lemma 5.13]. Their multiplicities
follow from a combination of Theorem 5.14 and Theorem 3.24 in [95], since the con-
struction in Sec. 3.4 of [95] yields idempotents eµ such that the commutative immage
of ±(eµ ) is equal to pµ /zµ . Explicitly, the eigenvalues of Sn ((1 ’ q)A) are i≥1 (1 ’ q µi ),
where µ = (µ1 , µ2 , . . . ) varies through all partitions of n, with corresponding multiplic-
ities n!/zµ , the number of permutations of cycle type µ, i.e., zµ = 1m1 m1! 2m2 m2! · · · ,
where mi is the number of occurences of i in the partition µ, i = 1, 2, . . . . Hence, the
eigenvalues of Kn (q) are (q; q)n/ i≥1 (1 ’ q µi ), with the same multiplicities.
Knowing all the eigenvalues of Kn (q) and their multiplicities explicitly, it is now not
extremely di¬cult to form the product of all these and, after a short calculation, recover
the right-hand side of (3.64).
Acknowledgements
I wish to thank an anonymous referee, Joris Van der Jeugt, Bernard Leclerc, Madan
Lal Mehta, Alf van der Poorten, Volker Strehl, Jean-Yves Thibon, Alexander Varchenko,
and especially Alain Lascoux, for the many useful comments and discussions which
helped to improve the contents of this paper considerably.
60 C. KRATTENTHALER

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