Library / Lie Groups and Lie Algebras / Chapter III / § 3
Proposition 43
Tag 0ABC, permanent. Cite it as [Bourbaki, Tag 0ABC].
Lie Groups and Lie Algebras, chapter III, § 3. Passage from a Lie group to its Lie algebra
Let t , u t, u t , u T ( ∞ ) ( G ) \mathscr{T}^{(\infty)}(G) T ( ∞ ) ( G ) ∑ i = 1 n t i ⊗ t i ′ \sum_{i=1}^n t_i \otimes t'_i ∑ i = 1 n t i ⊗ t i ′ t t t
t ⊤ u = ∑ i = 1 n t i ∗ u ∗ t i ′ ∨ .
t \top u = \sum_{i=1}^n t_i * u * {t'_i}^\vee.
t ⊤ u = i = 1 ∑ n t i ∗ u ∗ t i ′ ∨ . By definition, t ⊤ u t \top u t ⊤ u t ⊗ u t \otimes u t ⊗ u ( 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
α : ( g , g ′ ) ↦ ( g , g , g ′ ) of G × G into G × G × G β : ( g , g ′ , g ′ ′ ) ↦ ( g , g ′ − 1 , g ′ ′ ) of G × G × G into G × G × G γ : ( g , g ′ , g ′ ′ ) ↦ g g ′ ′ g ′ of G × G × G into G .
\begin{aligned}
\alpha & : (g, g') \mapsto (g, g, g') & \text{of } G \times G \text{ into } G \times G \times G \\
\beta & : (g, g', g'') \mapsto (g, {g'}^{-1}, g'') & \text{of } G \times G \times G \text{ into } G \times G \times G \\
\gamma & : (g, g', g'') \mapsto gg''g' & \text{of } G \times G \times G \text{ into } G.
\end{aligned}
α β γ : ( g , g ′ ) ↦ ( g , g , g ′ ) : ( g , g ′ , g ′′ ) ↦ ( g , g ′ − 1 , g ′′ ) : ( g , g ′ , g ′′ ) ↦ g g ′′ g ′ of G × G into G × G × G of G × G × G into G × G × G of G × G × G into G . On the other hand:
α ∗ ( t ⊗ u ) = ∑ i = 1 n ( t i ⊗ t i ′ ) ⊗ u = ∑ i = 1 n t i ⊗ t i ′ ⊗ u β ∗ ( ∑ i = 1 n t i ⊗ t i ′ ⊗ u ) = ∑ i = 1 n t i ⊗ t i ′ ∨ ⊗ u γ ∗ ( ∑ i = 1 n t i ⊗ t i ′ ∨ ⊗ u ) = ∑ i = 1 n t i ∗ u ∗ t i ′ ∨ .
\begin{aligned}
\alpha_*(t \otimes u) &= \sum_{i=1}^n (t_i \otimes t'_i) \otimes u = \sum_{i=1}^n t_i \otimes t'_i \otimes u \\
\beta_* \left( \sum_{i=1}^n t_i \otimes t'_i \otimes u \right) &= \sum_{i=1}^n t_i \otimes {t'_i}^\vee \otimes u \\
\gamma_* \left( \sum_{i=1}^n t_i \otimes {t'_i}^\vee \otimes u \right) &= \sum_{i=1}^n t_i * u * {t'_i}^\vee.
\end{aligned}
α ∗ ( t ⊗ u ) β ∗ ( i = 1 ∑ n t i ⊗ t i ′ ⊗ u ) γ ∗ ( i = 1 ∑ n t i ⊗ t i ′ ∨ ⊗ u ) = i = 1 ∑ n ( t i ⊗ t i ′ ) ⊗ u = i = 1 ∑ n t i ⊗ t i ′ ⊗ u = i = 1 ∑ n t i ⊗ t i ′ ∨ ⊗ u = i = 1 ∑ n t i ∗ u ∗ t i ′ ∨ . Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.
Cited by
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Cites
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Transcribed from the printed volume. Not translated.