Let ωg,ωt,ωG/T be invariant differential forms on g,t, G/T, respectively, of respective degrees n,r,n−r. If ωg=ωG/T∩ωt, we have
ψ∗ωg=ωG/T∧πgωt
where πg is the function on t defined by πg(x)=∏α∈R(G,T)δ(α)(x).
Denote by ωG (resp. ωT) the invariant differential form of maximum degree on G (resp. T) that coincides with ωg (resp. ωt) at the origin. Consider the commutative diagram
(G/T)×t⟶ψg
(Id,expT) expG.
(G/T)×T⟶f G
In view of Prop. 1 of no. 2 and the relation expT∗ωT=ωt, we deduce the equality
ψ∗expG∗ωG=ωG/T∧δgωt.
By Prop. 2,ψ∗expG∗ωG=(ψ∗λg)ψ∗ωg. Since the function λg is invariant under Ad(G), we have (ψ∗λg)(g,x)=λg(x)=πg(x)δg(x) for g∈G/T,x∈t.
It follows that the forms ψ∗ωG(g,x) and ωG/T(g)∧πg(x)ωt(x) coincide where δg(x) is non-zero, that is on the dense open subset (G/T)×tr; thus, they are equal, hence the proposition.
Choose invariant differential forms ωG on G and ωT on T, of maximum degree, such that ∣ωG∣=dg and ∣ωT∣=dt; denote by ωg (resp. ωt) the translation-invariant differential form on g (resp. t) that coincides with ωG(e) (resp. ωT(e)) at the origin, and dz (resp. dx) the Haar measure ∣ωg∣ (resp. ∣ωt∣). Reasoning as in no. 2, mutatis mutandis, gives the following proposition:
Bourbaki sets the proof directly after the statement, and the corpus does not separate them, so what is above is the statement and its proof together.
Nothing.