Báo cáo toán học: "On the Sn-modules generated by partitions of a given shape"

Tuyển tập các báo cáo nghiên cứu khoa học về toán học trên tạp chí toán học quốc tế đề tài: On the Sn-modules generated by partitions of a given shape. | On the Sn-modules generated by partitions of a given shape Daniel Kane1 and Steven Sivek2 1 Department of Mathematics Harvard University dankane@ 2Department of Mathematics Massachusetts Institute of Technology ssivek@ Submitted May 12 2008 Accepted Aug 24 2008 Published Aug 31 2008 Mathematics Subject Classification 05E10 Abstract Given a Young diagram A and the set Hx of partitions of 1 2 . A of shape A we analyze a particular S x -module homomorphism QHx QHx to show that QHx is a submodule of QHx whenever A is a hook n 1 1 . 1 with m rows n m or any diagram with two rows. 1 Introduction Let A A1 . An be a Young diagram and let Hx be the set of row tabloids of shape A that is Hx is the set of all ways to fill the diagram with the numbers 1 2 . IAI V Aị such that the order within a given row does not matter and neither does the order of two rows of the same size. Let Vx denote the set of column tabloids of shape A which are defined analogously and note that Vx Hx canonically. In particular Hx and Vx are the sets of partitions of 1 2 . IA g of shape A and A0 respectively. Each of these sets has an obvious S x action obtained by permuting the numbers which fill the tabloids. For any two tabloids r 2 Hx and A 2 Vx we define an orthogonality relation r A to be satisfied precisely when r c 1 for every row r of r and column c of A. We define a homomorphism of S x -modules QVx QHx in terms of a matrix Kx whose rows and columns are indexed by Hx and Vx respectively and whose entries K A are equal to 1 whenever r A and 0 otherwise. Then Pylyavskyy 5 conjectured that Kx has full rank modifying a false conjecture of Stanley 7 . This is of particular interest due to work of Black and List 1 in which they introduced this homomorphism in the case where A is an n X m rectangle and showed that the conjecture would imply Foulkes plethysm conjecture as follows THE ELECTRONIC JOURNAL OF COMBINATORICS 15 2008 R111 1 Proposition Black and List . If Knxm

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