Let ωG,ωG/T and ωT be invariant differential forms on G,G/T and T, respectively, of respective degrees n,n−r and r. If ωG= ωG/T∩ωT, then
f∗(ωG)=ωG/T∧δGωT
Clearly we can assume that ωG/T and ωT are non-zero; then the differential form (u,t)→ωG/T(u)∧ωT(t) on (G/T)×T is of degree n and everywhere non-zero; hence there exists a numerical function δ on (G/T)×T such that
f∗(ωG)(u,t)=δ(u,t)ωG/T(u)∧ωT(t)
Observe now that, for h∈G,u∈G/T,t∈ T, we have f(h.u,t)= (Int h)f(u,t); since ωG is invariant under inner automorphisms, it follows immediately that δ(h.u,t)=δ(u,t), so δ(u,t)=δ(e,t).
Denote by p:g→g/t the quotient map and by φ:g/t→g the map defined by
φ(p(X))= (Ad t−1)X−X for X∈g; recall (§5, no. 4, Lemma 4) that the tangent map
T(e,t)(f):Te(G/T)×Tt(T)→Tt(G)
takes (z,tH) to t(φ(z)+H) for z∈g/t,H∈t.
Let z1,...,zn−r be elements of g/t,H1,...,Hr elements of t. Then
f∗ωG(e,t)(z1,...,zn−r,tH1,...,tHr)
=ωG(t)(tφ(z1),...,tφ(zn−r),tH1,...,tHr) by the calculation of T(e,t)(f)
=ωG(e)(φ(z1),...,φ(zn−r),H1,...,Hr) since ωG is invariant
=ωG/T(e)(pφ(z1),...,pφ(zn−r)).ωT(e)(H1,...,Hr) (no.1, Lemma 1)
= det(pφ )ωG/T(e)(z1,...,zn−r).ωT(e)(H1,...,Hr)
=δG(t)ωG/T(e)(z1,...,zn−r).ωT(t)(tH1,...,tHr)
since ωT is invariant
=δG(t)(ωG/T∧ωT)(e,t)(z1,...,zn−r,tH1,...,tHr)
so f∗ωG(e,t)=δG(t)(ωG/T∧ωT)(e,t); thus, δ(e,t)=δG(t), hence the proposition.
Give the manifolds G,T and G/T the orientations defined by the forms ωG,ωT and ωG/T, respectively. These forms define invariant measures on G,T and G/T (Chap. III, §3, no. 16, Props. 55 and 56), also denoted by ωG,ωT and ωG/T.
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.