Bourbaki

Lemma 3

Tag 01F0, permanent. Cite it as [Bourbaki, Tag 01F0].

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

The inverse image on (G/T)×Tr(G/T)\times T_r of the measure dgdg on GrG_r under the local homeomorphism fr(f_r(Integration, Chap. V, §6, no. 6) is the measure μ⊗δGdt\mu\otimes \delta_Gdt, where μ\mu is the unique G-invariant measure on G/TG/T of total mass 1.

Choose an invariant differential form ωT\omega_T (resp. ωG/T\omega_{G/T}) on T (resp. G/TG/T) of maximal degree, such that the measure defined by ωT\omega_T (resp. ωG/T\omega_{G/T}) is equal to dtdt (resp. μ\mu). Put ωG=ωG/T∩ωT\omega_G=\omega_{G/T}\cap \omega_T. Lemma 2 implies that the measure defined by ωG\omega_G is equal to dgdg. Let U be an open subset of (G/T)×Tr(G/T)\times T_r such that frf_r induces an isomorphism from U to an open subset V of GrG_r. Let φ\varphi be a continuous function with compact support in V; denote also by φ\varphi the extension of φ\varphi to GrG_r which vanishes outside V. We have

∫Vφdg=∫VφωG=∫U(φ∘fr)fr∗(ωG) \int_V\varphi dg=\int_V\varphi \omega_G=\int_U(\varphi \circ f_r)f_r^*(\omega_G)

=∫U(φ∘fr)ωG/T∧δGωT=\int_U(\varphi \circ f_r)\omega_{G/T}\wedge \delta_G\omega_T (Prop.1.1)

=∫U(φ∘fr)dμ.δGdt =\int_U(\varphi \circ f_r)d\mu.\delta_Gdt

hence the lemma.

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

Cites

  • Integration, Chap. V, §6, no. 6 not in this corpus: INT V
  • Lemma 2
Transcribed from the printed volume. Not translated.