Proposition 1

Tag 0A8Y, permanent. Cite it as [Bourbaki, Tag 0A8Y].

Lie Groups and Lie Algebras, chapter III, § 2. Group of tangent vectors to a Lie group

Let X be a manifold of class C^r and m a law of composition of class C^r on X. If m is associative (resp. commutative), then T(m) is associative (resp. commutative).

If m is associative, then $m \circ (m \times \mathrm{Id}_X) = m \circ (\mathrm{Id}_X \times m)$ , whence

T(m) \circ (T(m) \times \mathrm{Id}_{T(X)}) = T(m) \circ (\mathrm{Id}_{T(X)} \times T(m))

and hence $T(m)$ is associative. Let $s$ be the mapping $(x, y) \mapsto (y, x)$ of $X \times X$ into $X \times X$ . If $m$ is commutative, then $m \circ s = m$ and hence

T(m) \circ T(s) = T(m).

But $T(s)$ is the mapping $(u, v) \mapsto (v, u)$ of $T(X) \times T(X)$ into $T(X) \times T(X)$ . Hence $T(m)$ is commutative.

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

Cited by

Corollary, § 2
Proposition 3, § 2

Cites

Nothing.