Bourbaki

Proposition 20

Tag 0A9Z, permanent. Cite it as [Bourbaki, Tag 0A9Z].

Lie Groups and Lie Algebras, chapter III, § 3. Passage from a Lie group to its Lie algebra

Let G, G' be Lie groups, X a manifold of class CrC^r and (g,x)↦gx(g, x) \mapsto gx (resp. (x,g′)↦xg′(x, g') \mapsto xg') a law of left (resp. right) operation of class CrC^r of G (resp. G') on X. Suppose that (gx)g′=g(xg′)(gx)g' = g(xg') for all x∈X,g∈G,g′∈G′x \in X, g \in G, g' \in G'. Let t∈T(s)(G)t \in \mathscr{T}^{(s)}(G), t′∈T(s′)(G′)t' \in \mathscr{T}^{(s')}(G') and f:X→Ff : X \to F be a function of class CrC^r such that s+s′≤rs + s' \leq r. Then

(t∗f)∗t′=t∗(f∗t′). (t * f) * t' = t * (f * t').
For all x∈Xx \in X,
⟨εx,(t∗f)∗t′⟩=⟨εx∗t′∨,t∗f⟩by (8)=⟨t∨∗(εx∗t′∨),f⟩(Proposition 17)=⟨t∨∗εx,f∗t′⟩(Proposition 2 and (11))=⟨εx,t∗(f∗t′)⟩by (5). \langle \varepsilon_x, (t * f) * t' \rangle = \langle \varepsilon_x * {t'}^\vee, t * f \rangle \quad \text{by (8)} \\ = \langle t^\vee * (\varepsilon_x * {t'}^\vee), f \rangle \quad \text{(Proposition 17)} \\ = \langle t^\vee * \varepsilon_x, f * t' \rangle \quad \text{(Proposition 2 and (11))} \\ = \langle \varepsilon_x, t * (f * t') \rangle \quad \text{by (5).}

In particular, consider G as operating on itself by left and right translations. If f:G→Ff : G \to F is a function of class CrC^r on G and t∈T(s)(G)t \in \mathscr{T}^{(s)}(G) (with s≤rs \leq r), t∗ft * f and f∗tf * t are, if s<∞s < \infty, functions of class Cr−sC^{r-s} on G. Further, let t′∈T(s′)(G)t' \in \mathscr{T}^{(s')}(G), with s+s′≤rs + s' \leq r. Then

(t∗f)∗t′=t∗(f∗t′). (t * f) * t' = t * (f * t').
In particular, C∞(G)\mathscr{C}^\infty(G) is a (T(∞)(G),T(∞)(G))(\mathscr{T}^{(\infty)}(G), \mathscr{T}^{(\infty)}(G))-bimodule. Formulae (5) and (8) admit as special cases
⟨t,f⟩=⟨εe,t∨∗f⟩=⟨εe,f∗t∨⟩. \langle t, f \rangle = \langle \varepsilon_e, t^\vee * f \rangle = \langle \varepsilon_e, f * t^\vee \rangle.

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

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