Proposition 6

Tag 0A9G, permanent. Cite it as [Bourbaki, Tag 0A9G].

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

Let $G, H$ be Lie groups and $\phi$ a morphism of $G$ into $H$ . For $t, t'$ in $\mathscr{T}^{(\infty)}(G)$ , $\phi_*(t * t') = \phi_*(t) * \phi_*(t')$ .

Consider the diagram

\begin{array}{ccc} G \times G & \xrightarrow{m} & G \\ \phi \times \phi \downarrow & & \downarrow \phi \\ H \times H & \xrightarrow{n} & H \end{array}

where

m(g, g') = gg'

n(h, h') = hh'

. This diagram is commutative. Hence

\begin{align*} \phi_*(t * t') &= \phi_*(m_*(t \otimes t')) = n_*((\phi \times \phi)_*(t \otimes t')) \\ &= n_*(\phi_*(t) \otimes \phi_*(t')) = \phi_*(t) * \phi_*(t'). \end{align*}

The Lie groups $G$ and $G^\vee$ have the same underlying manifold and hence the vector spaces $\mathscr{T}^{(\infty)}(G)$ and $\mathscr{T}^{(\infty)}(G^\vee)$ are the same. Let $\theta$ be the mapping $g \mapsto g^{-1}$ , which is an isomorphism of the Lie group $G$ onto the Lie group $G^\vee$ . Then $\theta^*$ is an automorphism of the vector space $\mathscr{T}^{(\infty)}(G)$ , which automorphism we denote by $t \mapsto t^\vee$ . Then $(\varepsilon_g)^\vee = \varepsilon_{g^{-1}}$ . If $t \in T_e(G)$ , then

t^\vee = -t \quad (§ 2, \text{Proposition 2}).

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

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Proposition 2