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

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

for $x \in m$; as

it follows immediately that

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

We write $A(X) = A, A^n(X) = A^n, \hat{A}(X) = \hat{A}, L^n(X) = L^n, \hat{L}(X) = \hat{L}, \Gamma(X) = \Gamma$. Let B be the algebra $A \otimes A$ with the graduation of type $\mathbf{N}$ defined by $B^n = \sum_{i+j=n} A^i \otimes A^j$. Let $\hat{B} = \prod_{n \geq 0} B^n$ be the associated complete filtered algebra (Commutative Algebra, Chapter III, § 2, no. 12, Example 1). The coproduct $c : A \to A \otimes A$ defined in § 3, no. 1, Corollary 1 to Theorem 1 is graded of degree 0 and hence extends by continuity to a homomorphism $\hat{c} : \hat{A} \to \hat{B}$ given by

We also define continuous homomorphisms $\delta'$ and $\delta''$ of $\hat{A}$ into $\hat{B}$ by

By Corollary 2 to Theorem 1 of § 3, no. 1, $L^n$ is the set of $a_n \in A^n$ such that $c(a_n) = a_n \otimes 1 + 1 \otimes a_n$. It follows that $\hat{L}$ is the set of $a \in \hat{A}$ such that

Let $\Delta$ be the set of $b \in \hat{A}$ of constant term equal to 1 and satisfying the relation

in other words, the set of $b = \sum_{n \geq 0} b_n$ such that $b_n \in \mathbf{A}^n$ for all $n \geq 0$, $b_0 = 1$ and $c(b_n) = \sum_{i+j=n} b_i \otimes b_j$ for $n \geq 0$. The latter characterization shows that $\Delta$ is a closed subset of $\Gamma$; as $\hat{\epsilon}$, $\delta'$ and $\delta''$ are ring homomorphisms and every element of $\delta'(\hat{\mathbf{A}})$ commutes with every element of $\delta''(\hat{\mathbf{A}})$, the restrictions to $\Gamma$ of the mappings $c$ and $\delta'\delta''$ are group homomorphisms and $\Delta$ is a subgroup of $\Gamma$.

By Proposition 1 of no. 1, the exponential mapping of $\hat{\mathbf{A}}$ is a bijection of the set $\hat{\mathbf{A}}^+$ of elements of $\hat{\mathbf{A}}$ with no constant term onto $\Gamma$. Let $a \in \hat{\mathbf{A}}^+$ and $b = \exp a$. As $\hat{\epsilon}$ is a continuous ring homomorphism,

The relations

are proved similarly and, as $\delta'(a)$ commutes with $\delta''(a)$ (no. 1, Remark 1),

Therefore $a$ satisfies (3) if and only if $b$ satisfies (4), which proves the theorem.

Every Lie polynomial of degree $n \geqslant 2$ is a finite sum of terms of the form $[Q, R]$ where $Q$ and $R$ are of degree $< n$ and the sum their degrees is equal to $n$ (§ 2, no. 7, Proposition 7). The lemma follows by induction on $n$.

A Lie formal power series† (with coefficients in $K$) in the indeterminates $(T_i)_{i \in I}$ is any element of the Lie algebra $\hat{L}((T_i)_{i \in I}) = \hat{L}(I)$. Such an element $u$ can be written uniquely as the sum of a summable family $(u_\nu)_{\nu \in \mathbf{N}^{(I)}}$ where $u_\nu \in L^\nu(I)$.

Suppose that $I$ is finite. Let $g$ be a complete Hausdorff filtered Lie algebra such that $g = \bigcup_{\alpha > 0} g_\alpha$; let $t = (t_i)_{i \in I}$ be a family of elements of $g$.

There exists $\alpha > 0$ such that $t_i \in g_\alpha$ for all $i \in I$; hence $f_t(L^\nu(I)) \subset g_{|\nu| \wedge \alpha}$ for all $\nu$ (Lemma 1), which implies the continuity of $\hat{f}_t$.

If $u \in \hat{L}(I)$, we write $u((t_i)) = \hat{f}_t(u)$. In particular, taking $g = \hat{L}(I)$, $u = u((T_i))$; in the general case, $u((t_i))$ is called the result of substituting the $t_i$ for the $T_i$ in the Lie formal power series $u((T_i))$. If $u = \sum_{\nu \in \mathbf{N}^{(I)}} u_\nu$, where $u_\nu \in L^\nu(X)$, the family $(u_\nu((t_i)))_{\nu \in \mathbf{N}^{(I)}}$ is summable and

Let $\sigma$ be a continuous homomorphism of $g$ into a complete Hausdorff filtered Lie algebra $g'$ such that $g' = \bigcup_{\alpha > 0} g'_\alpha$. For every finite family $t = (t_i)_{i \in I}$ of elements of $g$ and all $u \in \hat{L}(I)$,

for the homomorphism $\sigma \circ \hat{f}_t$ is continuous and maps $T_i$ to $\sigma(t_i)$ for $i \in I$.

Let $u = (u_j)_{j \in J}$ be a finite family of elements of $\hat{L}(I)$ and let $v \in \hat{L}(J)$; by substituting the $u_j$ for the $T_j$ in $v$, we obtain an element $w = v((u_j)_{j \in J})$ of $\hat{L}(I)$ denoted by $v \circ u$. Then

for every finite family $t = (t_i)_{i \in I}$ of elements of $g$, as is seen by operating with the continuous homomorphism $\hat{f}_t$ on the equation $w = v((u_j)_{j \in J})$.

Let $u = \sum_{\nu \in \mathbf{N}^{I}} u_\nu \in \hat{L}(I)$, where $u_\nu \in L^\nu(I)$. The mapping $\tilde{u}$ $(t_i) \mapsto u((t_i))$ of $g^I$ into $g$ is continuous: for in each of the open sets $g_\alpha$ with $\alpha > 0$ the family of $\tilde{u}_\alpha$ is uniformly summable and it suffices to prove that each $\tilde{u}_\nu$ is continuous, which is immediate by induction on $|\nu|$.

† A Lie formal power series is not in general a formal power series in the sense of Algebra, Chapter IV, § 6.

In $\hat{A}_\mathbf{Q}(\{U, V\})$, $\exp U \cdot \exp V = 1 + W$, where $W = \sum_{r+s \geq 1} \frac{U^r V^s}{r! s!}$, whence

The linear mapping $P_n$, defined by $P_n(x_1, \ldots, x_n) = \frac{1}{n} \left( \prod_{i=1}^{n-1} (\mathrm{ad}\, x_i) \right)(x_n)$ for $n \geq 1$ and $x_1, \ldots, x_n$ in $\{U, V\}$, is a projector of $A^n_\mathbf{Q}(\{U, V\})$ onto $L^n_\mathbf{Q}(\{U, V\})$ (§ 3, no. 2, Corollary to Proposition 1); as $H_{r,s}$ belongs to $L^{r+s}_\mathbf{Q}(\{U, V\})$, $H_{r,s} = P_{r+s}(H_{r,s})$. Now

when $s_m \geqslant 1$ and

when $r_m \geqslant 1$ and $s_m = 0$. Moreover, obviously $(\mathrm{ad}\ t)^{p-1}.t = 0$ if $p \geqslant 2$ and $(\mathrm{ad}\ t)^0.t = t$. It follows that the two sides of (12) are zero when $s_m \geqslant 2$ and those of (13) are zero when $r_m \geqslant 2$. The theorem then follows since $H'_{r,s}$ is the sum of the terms of type (12) and $H''_{r,s}$ is the sum of the terms of type (13).

Let $a, b$ be in $m$; there exists $\alpha > 0$ such that $a \in A_\alpha$ and $b \in A_\alpha$. Then there exists a continuous homomorphism $\theta$ of the Magnus algebra $\hat{A}(\{U, V\})$ into $A$ mapping $U$ to $a$ and $V$ to $b$ (§ 5, no. 1, Proposition 1).

The restriction of $\theta$ to $\hat{L}(\{U, V\})$ is a continuous homomorphism of Lie algebras of $L(\{U, V\})$ into $m$ which maps $U$ (resp. $V$) to $a$ (resp. $b$). By formula (6) of no. 3, therefore $\theta(H) = H(a, b)$. It then suffices to apply the continuous homomorphism $\theta$ to the two sides of the relation

taking account of Remark 3 of no. 1.

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

for $a, b, c$ in $g$. 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: This establishes relations (18) to (20).

We write $K(U, V) = H(U + V, -U)$, $K_1(U, V) = \sum_{n \geq 0} K_{n,1}(U, V)$. We denote by L (resp. R) left (resp. right) multiplication by $U$ on $\hat{A}(\{U, V\})$.

We can write

and therefore We now calculate modulo the ideal $\sum_{m \geq 0} \sum_{n \geq 2} A^{m,n}(\{U, V\})$ of $A(\{U, V\})$. For all $n \geq 1$, whence Therefore summing over $n$.

On the other hand, $K_1(U, V) \equiv K(U, V)$ and $e^{K_1(U, V)} \equiv 1 + K_1(U, V)$ and hence

by Proposition 3. We deduce that It then suffices to apply the Remark of § 2, no. 11.