Proposition 7

Tag 0A9I, permanent. Cite it as [Bourbaki, Tag 0A9I].

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

Let G be a Lie group. Let t, t' be in $\mathscr{T}^{(\infty)}(G)$ .

(i) The product $t * t'$ calculated relative to $G^\vee$ is equal to the product $t' * t$ calculated relative to G.

(ii) $(t * t')^\vee = {t'}^\vee * t^\vee$ .

Consider the diagram

\begin{array}{ccc} (G_1 \times G_2) \times (G_1 \times G_2) & \xrightarrow{m} & G_1 \times G_2 \\ \downarrow n & & \uparrow p_1 \times p_2 \\ (G_1 \times G_1) \times (G_2 \times G_2) \end{array}

where $s(g, g') = (g', g)$ , $m(g, g') = gg'$ , $n(g, g') = g'g$ for all $g, g'$ in G. This diagram is commutative. Hence $n_*(t \otimes t') = m_*(s_*(t \otimes t')) = m_*(t' \otimes t)$ . This equality is precisely (i). Assertion (ii) follows from (i) and Proposition 6.

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

Cited by

Cites

Proposition 6