On the algebraic set of singular elements in a complex simple Lie algebra

نویسندگان

  • BERTRAM KOSTANT
  • NOLAN WALLACH
چکیده

Let G be a complex simple Lie group and let g = LieG. Let S(g) be the G-module of polynomial functions on g and let Sing g be the closed algebraic cone of singular elements in g. Let L ⊂ S(g) be the (graded) ideal defining Sing g and let 2r be the dimension of a G-orbit of a regular element in g. Then Lk = 0 for any k < r. On the other hand, there exists a remarkable G-module M ⊂ Lr which already defines Sing g. The main results of this paper are a determination of the structure of M . 0. Introduction 0.1. Let G be a complex simple Lie group and let g = LieG. Let = rank g. Then in superscript centralizer notation one has dim g ≥ for any x ∈ g. An element x ∈ g is called regular (resp. singular) if dim g = (resp. > ). Let Reg g be the set of all regular elements in g and let Sing g, its complement in g, be the set of all singular elements in g. Then one knows that Reg g is a nonempty Zariski open subset of g and hence Sing g is a closed proper algebraic subset of g. Let S(g) (resp. ∧ g) be the symmetric (resp. exterior) algebra over g. Both algebras are graded and are G-modules by extension of the adjoint representation. Let B be the natural extension of the Killing form to S(g) and ∧ g. The inner product it induces on u and v in either S(g) or ∧ g is denoted by (u, v). The use of B permits an identification of S(g) with the algebra of polynomial functions on g. Since Sing g is clearly a cone the ideal, L, of all f ∈ S(g) which vanish on Sing g is graded. Let n = dim g and let r = (n− )/2. One knows that n− is even so that r ∈ Z+. It is easy to show that L = 0, for all k < r. (0.1) 1 The purpose of this paper is to define and study a rather remarkable G-submodule

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تاریخ انتشار 2010