AM-finite functions

Let M be a standard Levi subgroup of G.

Remark

Let Hom(AM,) be the set of characters of AM, i.e. of continuous homomorphisms of AM into. The restriction map of M to AM defines a map

(1) We can use this to prove surjectivity by giving a description of Hom(AM) analogous to the description of XM (see I.1.4). As AMM1 is of finite index in M, the kernel of (1) is finite. It contains since AM⊂ZM. We will call it XM(AM). We obtain an isomorphism

Suppose first that k is a number field. Let (AM) be the enveloping algebra of the (complex) Lie algebra of the real group AM. We have

Thus (4M) is identified, by a map which we will denote by z ↦, with the polynomial algebra over the complex space, itself isomorphic to XM and even to XM/XM(AM), since XM(AM) = {0}. Suppose now that k is a function field. Let (AM) be the convolution algebra of functions with compact support on AM. We associate with z ∊ (AM) its Fourier-Mellin transform, which is the function on XM/XM(AM) defined by

Fix a basis (ai=1,…,n of the ℤ-module ZM. Then (AM) can be identified via z ↦ with the space of polynomials in the variables for i = 1,…,n.