Bourbaki

Proposition 5

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

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

Assume that the group H is compact and denote by dhdh the Haar measure on H of total mass 1. Let ss be a section of E of class CrC^r. Denote by s♯s^{\sharp} the vector integral ∫Hhsdh\int_H^hs dh. Then s♯s^{\sharp} is a section of E of class CrC^r, invariant under H; for x∈Xx\in X, we have s♯(x)=∫Hhs(h−1x)dh∈Exs^{\sharp}(x) =\int_Hhs(h^{-1}x)dh\in E_x. The endomorphism s→s♯s \rightarrow s^{\sharp} of Sr(X;E)\mathscr{S}^r(X; E) is a projection onto the subspace of H-invariant sections.

Consider the map h→hsh \rightarrow^hs from H to Sr(X;E)\mathscr{S}^r(X; E); it is continuous by Lemma 4. Since the space Sr(X;E)\mathscr{S}^r(X; E) is separated and complete, the integral s♯=s^{\sharp}= ∫Hhsdh\int_H^hs dh belongs to Sr(X;E)(\mathscr{S}^r(X; E) (Integration, Chap. III, §3, no. 3, Cor. 2). The linear map s→s(x)s \rightarrow s(x) from Sr(X;E)\mathscr{S}^r(X; E) to ExE_x being continuous, we have s♯(x)=s^{\sharp}(x) = ∫Hhs(x)dh\int_H^hs(x)dh for all x∈Xx\in X. It is clear that s♯s^{\sharp} is invariant under H; if ss is an H-invariant section, we have s♯=ss^{\sharp}=s, hence the last assertion.

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

Nothing.

Cites

  • Integration, Chap. III, §3, no. 3, Cor. 2 not in this corpus: INT III
  • Lemma 4
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