Remark 1

Tag 0A9Y, permanent. Cite it as [Bourbaki, Tag 0A9Y].

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

Let G be a Lie group, X a manifold of class $C^r$ and $(x, g) \mapsto xg$ a law of right operation of class $C^r$ of G on X. If $t \in \mathscr{T}^{(s)}(G)$ with $s \leq r$ and $f : X \to F$ is a function of class $C^r$ on X, we denote by $f * t$ the function on X defined by

\langle \varepsilon_x, f * t \rangle = \langle \varepsilon_x * t^\vee, f \rangle \\ = \langle \rho(x) * (t^\vee), f \rangle \\ = \langle t^\vee, f \circ \rho(x) \rangle \\ = \langle t, (f \circ \rho(x))^\vee \rangle.

In particular

(f * \varepsilon_g)(x) = f(xg^{-1})

that is

f * \varepsilon_g = \delta(g)^{-1} f.

Propositions 17, 18, 19 become, in the obvious notation,

\langle t', f * t \rangle = \langle t' * t^\vee, f \rangle \tag{11}

f * (t * t') = (f * t) * t' \tag{12}

(ff') * t = \sum_{i=1}^n (f * t_i)(f' * t_i') \tag{13}

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

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