Library / Lie Groups and Lie Algebras / Chapter III / § 3
Proposition 46
Tag 0ABJ, permanent. Cite it as [Bourbaki, Tag 0ABJ].
Lie Groups and Lie Algebras, chapter III, § 3. Passage from a Lie group to its Lie algebra
Let h ∈ G , x ∈ T h ( G ) h \in G, x \in T_h(G) h ∈ G , x ∈ T h ( G ) a ∈ L ( G ) a \in L(G) a ∈ L ( G ) ϕ \phi ϕ ( g , g ′ ) ↦ g g ′ g − 1 (g, g') \mapsto gg'g^{-1} ( g , g ′ ) ↦ g g ′ g − 1 G × G G \times G G × G G G G ( a , x ) ∈ T e ( G ) × T h ( G ) (a, x) \in T_e(G) \times T_h(G) ( a , x ) ∈ T e ( G ) × T h ( G ) T ( e , h ) ( ϕ ) T_{(e, h)}(\phi) T ( e , h ) ( ϕ ) y = x + h ( ( A d h − 1 ) a − a ) y = x + h((\mathrm{Ad}\,h^{-1})a - a) y = x + h (( Ad h − 1 ) a − a )
y = ( T ( e , h ) ϕ ) ( a ⊗ ε h + ε e ⊗ x ) = a ⊤ ε h + ε e ⊤ x = a ∗ ε h − ε h ∗ a + x = h ( ( A d h − 1 ) a ) − h a + x .
\begin{align*}
y &= (T_{(e, h)}\phi)(a \otimes \varepsilon_h + \varepsilon_e \otimes x) \\
&= a \top \varepsilon_h + \varepsilon_e \top x \\
&= a * \varepsilon_h - \varepsilon_h * a + x \\
&= h((\mathrm{Ad}\,h^{-1})a) - ha + x.
\end{align*}
y = ( T ( e , h ) ϕ ) ( a ⊗ ε h + ε e ⊗ x ) = a ⊤ ε h + ε e ⊤ x = a ∗ ε h − ε h ∗ a + x = h (( Ad h − 1 ) a ) − ha + x . Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.
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Transcribed from the printed volume. Not translated.