tailieunhanh - Báo cáo toán học: "The Action of the Symmetric Group on a Generalized Partition Semilattice"

Tuyển tập các báo cáo nghiên cứu khoa học trên tạp chí toán học quốc tế đề tài: The Action of the Symmetric Group on a Generalized Partition Semilattice. | The Action of the Symmetric Group on a Generalized Partition Semilattice Robert Gill Department of Mathematics and Statistics University of Minnesota Duluth 10 University Drive Duluth MN 55812 rgill@ Submitted February 7 1998 Accepted April 21 2000 Key Words hyperplane arrangement Mobius function homology induced character cyclic group AMS Subject Classification 1991 Primary 06A07 Secondary 05E25 52B30 20C30 11A05 Abstract Given an integer n 2 and a non-negative integer k consider all affine hyperplanes in Rn of the form Xi Xj r for i j 2 n and a non-negative integer r k. Let nn k be the poset whose elements are all nonempty intersections of these affine hyperplanes ordered by reverse inclusion. It is noted that nn 0 is isomorphic to the well-known partition lattice nn and in this paper we extend some of the results of nn by Hanlon and Stanley to nn k. Just as there is an action of the symmetric group n on nn there is also an action on nn fe which permutes the coordinates of each element. We consider the subposet nn k of elements that are fixed by some Ơ 2 n and find its Mobius function ụ ơ using the characteristic polynomial. This generalizes what Hanlon did in the case k 0. It then follows that 1 n-1 ơ nn k as a function of Ơ is the character of the action of n on the homology of nn k. Let Tn k be this character times the sign character. For Cn the cyclic group generated by an n-cycle Ơ of n we take its irreducible characters and induce 1 THE ELECTRONIC .JOURNAL OF COMBINATORICS 7 2000 R23 2 them up to n. Stanley showed that Tn 0 is just the induced character X ff where x ơ e dn. We generalize this by showing that for k 0 there exists a non-negative integer combination of the induced characters described here that equals n k and we find explicit formulas. In addition we show another way to prove that n k is a character without using homology by proving that the derived coefficients of certain induced characters of n are non-negative integers. 1 .