Proposition 1

Tag 0A61, permanent. Cite it as [Bourbaki, Tag 0A61].

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

The exponential mapping is a homeomorphism of $m$ onto $1 + m$ and the logarithmic mapping is the inverse homeomorphism.

For $x \in A_\alpha$ , $\frac{x^n}{n!} \in A_{n\alpha}$ . It follows that the series defining the exponential converges uniformly on each of the sets $A_\alpha$ for $\alpha > 0$ ; as $A_\alpha$ is open in $m$ and $m = \bigcup_{\alpha > 0} A_\alpha$ , the exponential mapping is continuous. It can be similarly shown that the logarithmic mapping is continuous.

Let $e$ and $l$ be the formal power series with no constant term

e(X) = \sum_{n \geq 1} \frac{X^n}{n!}, \quad l(X) = \sum_{n \geq 1} (-1)^{n-1} X^n / n.

We know (Algebra, Chapter IV, § 6, no. 9) that $e(l(X)) = l(e(X)) = X$ on $\hat{A}(\{X\}) = K[[X]]$ . By substitution (§ 5, no. 1), we deduce that

e(l(x)) = l(e(x)) = x

for $x \in m$ ; as

\exp x = e(x) + 1, \quad \log(1 + x) = l(x)

it follows immediately that

\log \exp x = x, \quad \exp \log(1 + x) = 1 + x

for $x$ in $m$ , whence the proposition.

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

Proposition 1

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