Proposition 4

Tag 0A6C, permanent. Cite it as [Bourbaki, Tag 0A6C].

Lie Groups and Lie Algebras, chapter II, § 6. The Hausdorff series

Let $g$ be a complete Hausdorff filtered Lie algebra such that $g = \bigcup_{\alpha > 0} g_\alpha$ . The mapping $(a, b) \mapsto H(a, b)$ is a group law on $g$ compatible with the topology on $g$ under which $0$ is the identity element and $-a$ is the inverse of $a$ for all $a \in g$ .

The mapping $(a, b) \mapsto H(a, b)$ of $g \times g$ into $g$ is continuous (no. 3); as the mapping $a \mapsto -a$ is obviously continuous, it suffices to prove the relations

\begin{align*} (18) \quad & H(H(a, b), c) = H(a, H(b, c)) \\ (19) \quad & H(a, -a) = 0 \\ (20) \quad & H(a, 0) = H(0, a) = a \end{align*}

for

a, b, c

g

. By formula (7) of no. 3, it suffices to prove these formulae when

a, b, c

are three indeterminates and

g = \hat{L}(\{a, b, c\})

. Now the restriction of the exponential mapping to

\hat{L}(\{a, b, c\})

is an injection into the Magnus algebra

\hat{A}(\{a, b, c\})

and by Proposition 3:

\begin{align*} \exp H(H(a, b), c) &= \exp H(a, b) . \exp c = \exp a . \exp b . \exp c \\ \exp H(a, H(b, c)) &= \exp a . \exp H(b, c) = \exp a . \exp b . \exp c \\ \exp H(a, -a) &= \exp a . \exp(-a) = \exp(a - a) = \exp 0 \\ \exp H(a, 0) &= \exp a . \exp 0 = \exp a \\ \exp H(0, a) &= \exp 0 . \exp a = \exp a. \end{align*}

This establishes relations (18) to (20).

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

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Cites

Proposition 3