Proposition 14

Suppose that $p \neq 0$ and that v is chosen such that $v(p) = 1$. Let G be a standard group and E(X) (resp. L(X)) the expansion of the exponential function of G (resp. the logarithmic function of G) as an integral series about 0.

(i) The domain of absolute convergence (Differentiable and Analytic Manifolds, R, 4.1.3) of E contains the set $\Delta$ of $x \in G$ such that $\omega(x) > \frac{1}{p-1}$. Let $E'$ denote the mapping defined on $\Delta$ by this series. Then $E'$ is an exponential mapping of G and is an isomorphism of the manifold $\Delta$ onto itself.

(ii) The domain of absolute convergence of L contains G. Let $L'$ denote the mapping on G defined by this series. Then $L'$ is the logarithmic mapping of G and the restriction of $L'$ to $\Delta$ is the inverse mapping of $E'$.

(iii) The mapping $E'$ is an isomorphism of $\Delta$, with the Hausdorff law, onto the subgroup $\Delta$ of G.

Using the notation of § 5, nos. 3 and 4, $E = \sum_{m \geq 1} \frac{\psi_{m,m}}{m!}$ ($§ 5$, no. 4, Proposition 3). As the coefficients $c_{\alpha \beta \gamma}$ belong to A, $\| \psi_{m,m} \| \leq 1$ (Differentiable and Analytic Manifolds, R, Appendix) $K^r$ is assumed to have the norm

By Chapter II, § 8, no. 1, Lemma 1, $v(m!) \leq \frac{m-1}{p-1}$. If $\omega(x) > \frac{1}{p-1}$, we see that $m \omega(x) - v(m!)$ tends to $+\infty$ with $m$, whence

which tends to 0 as $m$ tends to $+\infty$ and Therefore $\Delta$ is contained in the domain of absolute convergence of $E$ and $E'(\Delta) \subset \Delta$. Clearly $E'$ is an exponential mapping.

If $L_m$ denotes the homogeneous component of $L$ of degree $m$, Proposition 3 of § 5, no. 4, proves that each coefficient of $L_m$ is of the form

with $a_1, a_2, \ldots, a_m$ in $A$; but and Therefore, if $\omega(x) > 0$, $\|L_m\| \cdot \|x\|^m$ tends to 0 as $m$ tends to $+\infty$, so that $G$ is contained in the domain of absolute convergence of $L$. On the other hand, if $\omega(x) > \frac{m}{p-1}$, then $\omega(L_m(x)) > \frac{m}{p-1} - \frac{m-1}{p-1} = \frac{1}{p-1}$ for $m \geq 1$ and hence $L'(\Delta) \subset \Delta$.

As the formal power series $L(E(X))$ and $E(L(X))$ are equal to $X$, no. 4.1.5 of Differentiable and Analytic Manifolds, R, proves that

for $x \in \Delta$. Hence $E'$ is an isomorphism of the manifold $\Delta$ onto itself and the inverse isomorphism is the restriction of $L'$ to $\Delta$.

$L(X^{[n]}) = nL(X)$ for $n$ an integer $> 0$ (cf. § 5, no. 4). As $G$ is contained in the domain of absolute convergence of $L$ and $X^{[n]}$, therefore $L'(x^n) = nL'(x)$ for all $x \in G$. The relation $L'|_{\Delta} = {E'}^{-1}$ implies that $L'(x^n) = \log x^n$ for $n$ sufficiently large. Hence $L'(x) = \log x$. We have thus proved (i) and (ii).

Let $H = \sum_{r,s \geq 0} H_{r,s}$ be the Hausdorff formal power series and $h$ the Hausdorff function relative to $L(G)$. The domain of absolute convergence of $\tilde{H}$ contains $\Delta \times \Delta$ and $h$ is defined on $\Delta \times \Delta$ (Chapter II, § 8, Proposition 2). Then

for $x, y$ sufficiently close to 0 ($§ 4$, Theorem 4 (v)). Hence, in the notation of no. 3, Definition 1, the formal power series $F(E(X), E(Y))$ and $E(H(X, Y))$ are equal. Let $x, y$ be elements of $\Delta$. Then by Chapter II, § 8, formula (14). By Differentiable and Analytic Manifolds, R, 4.1.5, $E'(x)E'(y)$ is obtained by substituting $x$ for $X$ and $y$ for $Y$ in and $E'(h(x, y))$ is obtained by substituting $x$ for $X$ and $y$ for $Y$ in $E(H(X, Y))$. Hence $E'(x)E'(y) = E'(h(x, y))$.