Bourbaki

Lemma 1

Tag 0AA8, permanent. Cite it as [Bourbaki, Tag 0AA8].

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

Let X and X′X' be complete normable spaces, X0X_0 an open neighbourhood of 0 in X and f an analytic mapping of X0X_0 into X′X' such that f(0)=0f(0) = 0. Let f=f1+f2+f3+⋯f = f_1 + f_2 + f_3 + \cdots be the expansion of f as an integral series about 0, where fif_i is a homogeneous continuous-polynomial of degree i on X with values in X′X'. Let t be an element of TS2(X)^2(X), considered as a point distribution on X with support contained in \{0\}. Let t′=f∗(t)∈TS(X′)t' = f_*(t) \in TS(X'). The homogeneous component of t′t' of degree 1 is ⟨f2,t⟩\langle f_2, t \rangle.

Let t1′t'_1 be this component. Then, for every continuous linear mapping u of X′X' into a polynormed space,

u(t1′)=⟨t′,u⟩because u is continuous and linear u(t'_1) = \langle t', u \rangle \quad \text{because } u \text{ is continuous and linear}
=⟨t,u∘f⟩(Diff. & Anal. Man., R, 13.2.3) = \langle t, u \circ f \rangle \quad (\text{Diff. \& Anal. Man., R, 13.2.3})
=⟨t,u∘f2⟩because t∈TS2(X) = \langle t, u \circ f_2 \rangle \quad \text{because } t \in TS^2(X)
=u(⟨t,f2⟩)(Diff. & Anal. Man., R, 13.2.2), = u(\langle t, f_2 \rangle) \quad (\text{Diff. \& Anal. Man., R, 13.2.2}),

whence the lemma.

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

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Cites

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Transcribed from the printed volume. Not translated.