Remark

(1) If $x \in m, y \in m$ and $x$ and $y$ commute, then

since the family $\left( \frac{x^i}{i!} \cdot \frac{y^j}{j!} \right)_{i,j \in \mathbf{N}}$ is summable (cf. Algebra, Chapter IV, § 6, no. 9, Proposition 11).

(2) As the series $e$ and $l$ are without constant term and $A_\alpha$ is a closed ideal of $A$, $\exp A_\alpha \subset 1 + A_\alpha$ and $\log(1 + A_\alpha) \subset A_\alpha$ whence $\exp A_\alpha = 1 + A_\alpha$ and $\log(1 + A_\alpha) = A_\alpha$ for $\alpha > 0$.

(3) Let $B$ be a complete Hausdorff filtered unital associative algebra and $n = \bigcup_{\alpha > 0} B_\alpha$. Let $f$ be a continuous unital homomorphism of $A$ into $B$ such that $f(m) \subset n$. Then $f(\exp x) = \exp f(x)$ for $x \in m$ and $f(\log y) = \log f(y)$ for $y \in 1 + m$; we show for example the first of these formulae:

(4) Let E be a unital associative algebra. If a is a nilpotent element of E, the family $\left( \frac{a^n}{n!} \right)_{n \in \mathbf{N}}$ has finite support and we write $\exp a = \sum_{n \geq 0} a^n / n!$. An element b is unipotent if $b - 1$ is nilpotent; then we write

We deduce from the relations $e(l(X)) = l(e(X)) = X$ that the mapping $a \mapsto \exp a$ is a bijection of the set of nilpotent elements of E onto the set of unipotent elements of E and that $b \mapsto \log b$ is the inverse mapping.