Lemma 2

Tag 01EZ, permanent. Cite it as [Bourbaki, Tag 01EZ].

Lie Groups and Lie Algebras, chapter IX, § 6. Integration on compact Lie groups

If $\omega_G=\omega_{G/T}\cap \omega_T$ , then

\int_G\omega_G=\int_{G/T}\omega_{G/T}.\int_T\omega_T

Denote by $\pi$ the canonical morphism from G to $G/T$ . Let $g\in G$ , and let $t_1, . . . , t_{n-r}$ be elements of $T_{\pi(g)}(G/T)$ . Identify the fibre $\pi^{-1}(\pi (g)) =gT$ with T by the translation $\gamma (g)$ . The relation $\omega_G=\omega_{G/T}\cap \omega_T$ now implies the equality (Differentiable and Analytic Manifolds, Results, 11.4.5):

\omega_G\llcorner (t_1, . . . , t_{n-r}) = (\omega_{G/T}(t_1, . . . , t_{n-r}))\omega_T

Thus $\int_{\pi}\omega_G=(\int_T\omega_T)\omega_{G/T}($ Differentiable and Analytic Manifolds, Results, 11.4.6), and

\int_G\omega_G=\int_{G/}\int_{T\pi}\omega_G=\int_T\omega_T.\int_{G/T}\omega_{G/T}

(Differentiable and Analytic Manifolds, Results, 11.4.8).

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.

Cited by

Lemma 3, § 6

Cites

Nothing.