Affine group. For every X∈GL(n,K) and every x∈Kn, let (X,x) be the affine linear mapping ξ↦Xξ+x in Kn. The set of (X,x) is the affine group G of Kn (A, II, §9, No. 4). The set T of translations is a closed normal subgroup of G, canonically isomorphic to Kn; on the other hand, GL(n,K) is a closed subgroup of G, and G is the semi-direct product of GL(n,K) and T=Kn. One equips G with the (locally compact) topology for which G is the topological semi-direct product of GL(n,K) and T (GT, III, §2, No. 10). One has
(X,x)=(1,x)⋅(X,0). On the other hand, if
X∈GL(n,K) and
x∈T then, for every
ξ∈Kn,
(X,0)(1,x)(X,0)−1ξ=X(X−1ξ+x)=ξ+Xx=(1,Xx)ξ, therefore the automorphism
(1,x)↦(X,0)(1,x)(X,0)−1 of
T has modulus
mod(detX) (§1, No. 10, Prop. 15). In view of Example 1 and §2, No. 9,
Remark, the measure
(5)mod(detX)−n−1⋅(ij⨂dxij)⊗(i⨂dxi)(X=(xij), x=(xi)) is a left Haar measure on
G. On the other hand, by Prop. 14 of §2, No. 9,
ΔG((X,x))=ΔGL(n,K)(X)ΔKn(x)(moddetX)−1, or
(6)ΔG((X,x))=mod(detX−1).
Thus, a right Haar measure on G is given by
(moddetX)−n⋅(ij⨂dxij)⊗(i⨂dxi).
Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.
Nothing.