Lemma 2

Let $(g_i)_i$ be a family of elements of order $\geq 2$ in $\mathbf{A}[[\mathbf{I}]]$. When $\mathbf{I}$ is infinite, assume that $g_i$ tends to 0 along the filter of complements of finite subsets of $\mathbf{I}$. There exists one and only one automorphism T of the topological A*-algebra* $\mathbf{A}[[\mathbf{I}]]$ such that $T(X_i) = X_i + g_i$ for all $i \in \mathbf{I}$. *Further,

(23)

for each $u \in \mathbf{A}[[\mathbf{I}]]$.

The series $f_i = X_i + g_i$ has no constant term and when $\mathbf{I}$ is infinite, $f_i$ tends to 0 along the filter of complements of finite subsets of $\mathbf{I}$. Therefore (IV, p. 28, Prop. 4) there exists precisely one continuous endomorphism T of the A-algebra $\mathbf{A}[[\mathbf{I}]]$ such that $T(X_i) = f_i$ for all $i \in \mathbf{I}$. For each $\nu \in \mathbf{N}'$ we put

the relations $\omega(g_i) \geq 2$ imply $w(v_\nu) \geq |\nu| + 1$, and the relation (23) follows at once from this.

Let us show that $T$ is injective. Given $u \in \mathbf{A}[[\mathbf{I}]]$ such that $T(u) = 0$, by (23) we have $w(u) \geq w(u) + 1$, which is impossible if $u \neq 0$ because $w(u)$ would then be a positive integer.

For every formal series $v$ in $\mathbf{A}[[\mathbf{I}]]$ we denote by $H_n(v)$ its homogeneous component of degree $n$. Let us put $S_0(v) = H_0(v)$ and define the continuous mappings $S, : \mathbf{A}[[\mathbf{I}]] \to \mathbf{A}[[\mathbf{I}]]$ by the recursion equations

(24)

Put $S(v) = \sum_{n \geq 0} S_n(v)$; if $\nu \in \mathbf{N}^{(\mathbf{I})}$ and $n = |\nu|$, then the coefficient $S^\nu(v)$ of $X^\nu$ in $S(v)$ is equal to that of $X^\nu$ in $S,(\nu)$; since $S,$ is a continuous mapping, the mapping $S^\nu : \mathbf{A}[[\mathbf{I}]] \to \mathbf{A}$ is continuous. Hence, by the definition of the product topology on $\mathbf{A}[[\mathbf{I}]] = \mathbf{A}^{\mathbf{N}^{(\mathbf{I})}}$, the mapping $S : \mathbf{A}[[\mathbf{I}]] \to \mathbf{A}[[\mathbf{I}]]$ is continuous.

We shall prove the relation $T(S(v)) = v$ for all $v \in \mathbf{A}[[\mathbf{I}]]$ which will complete the proof of the lemma. Let $v \in \mathbf{A}[[\mathbf{I}]]$, $u_n = S_n(v)$ and $u = S(v)$. Let $n$ be a positive integer such that

(25)$_n$

We have $w(u - (u_0 + \ldots + u_{n-1})) \geq n$, whence

by (23). Now the recursion equation (24) shows that

By (26) the formal power series $v - T(u)$ and $v - T(u_0 + \ldots + u_{n-1}) - u_n$ have the same homogeneous component of degree $n$, and this component is zero, by (27). We thus have $w(v - T(u)) \geq n + 1$, that is $(25)_n$ implies $(25)_{n+1}$. Since $(25)_0$ clearly holds, we thus have $\omega(v - T(u)) \geq n$ for every integer $n \geq 0$, whence $v = T(u) = T(S(v))$, as was to be proved.

For the rest of this No. we shall, for any set $I$, denote by $A\{I\}$ the set of families $(f_i)_{i \in I}$ satisfying the following conditions: (i) for each $i \in I$, $f_i$ is an element of $A[[I]]$ without constant term; (ii) if $I$ is infinite, $f_i$ tends to 0 along the filter of complements of finite subsets of $I$.

The set $A\{I\}$ is a monoid for the composition law $(f, g) \mapsto f \circ g$, with $\{X_i\}_{i \in I}$ as unit element. The set of invertible elements of $A\{I\}$ is thus a group.

On the other hand, let $E$ be the monoid of all continuous unital endomorphisms of the $A$-algebra $A[[I]]$ leaving the ideal of all formal power series without constant term invariant. If $f \in A\{I\}$ and $g \in A[[I]]$, then the element $g(f)$ is defined. For fixed $f$, the mapping $g \mapsto g(f)$ of $A[[I]]$ into itself is an element $W_f$ of $E$. If $f_1, f_2 \in A\{I\}$ and $g \in A[[I]]$, we have, by formula (5) (IV, p. 30)

hence $f \mapsto W_f$ is a homomorphism of the monoid opposite to $A\{I\}$ into $E$. By Prop. 4 (IV, p. 28) this homomorphism is bijective.

Let $f = (f_i)_{i \in I} \in A\{I\}$ and let $\sum_{j \in I} \alpha_{ij} X_j$ be the homogeneous component of degree 1 of $f_i$. For any fixed $j$ in $I$ we have $\alpha_{ij} = 0$ except for a finite number of suffixes $i$, by hypothesis (ii) above. If $(\lambda_i) \in A^{(I)}$, we thus have $\left( \sum_{j \in I} \alpha_{ij} \lambda_j \right) \in A^{(I)}$.

We denote by $T_f$ the $A$-linear mapping $^1$

of $A^{(I)}$ into $A^{(I)}$. If $g \in A\{I\}$, it is easily verified that

$^1$ Sometimes $T_f$ is called the linear mapping tangent to $f$.