Proposition 58

Tag 0AC6, permanent. Cite it as [Bourbaki, Tag 0AC6].

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

Let $G$ and $H$ be two Lie groups, $M$ a manifold of class $C^r$ , $f$ a mapping of class $C^r$ of $M$ into $G$ and $h$ a morphism of $G$ into $H$ . Then

(h \circ f)^{-1}.d(h \circ f) = L(h) \circ (f^{-1}.df) = (h^{-1}.dh) \circ T(f).

For all $x \in M$ and $u \in T_x(M)$ ,

(h \circ f)^{-1}.d(h \circ f)(u) = ((h \circ f)(x))^{-1}.T(h \circ f)(u).

The latter expression is equal, on the one hand, to

T(h)(f(x)^{-1}.T(f)(u)) \quad (§ 2, \text{Proposition 5})

= T_e(h)((f^{-1}.df)(u))

and, on the other hand, to

h(f(x))^{-1}T(h)(T(f)u)

= (h^{-1}.dh)(T(f)u).

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

Cited by

Nothing.

Cites

Proposition 5