The condition is clearly necessary. Conversely, if it is satisfied, let $(V_n)$ be a fundamental system of neighbourhoods of $e$; if $U_n$ denotes the set of pairs $(x, y) \in G \times G$ such that $x^{-1}y \in V_n$, then the $U_n$ form a countable fundamental system of entourages of the left uniformity of $G$; since this uniformity is Hausdorff, it follows from § 2, no. 4, Theorem 1 that it is metrizable. Similarly for the right uniformity of $G$.

A topological group $G$ is said to be metrizable if its topology is metrizable. Proposition 1 then shows that its two uniformities are metrizable.

This result can be sharpened with the help of the following notion:

Suppose that the fundamental system $(V_n)$ of neighbourhoods of $e$ consists of symmetric neighbourhoods such that $V_{n+1}^3 \subset V_n$ for each $n$. Then the corresponding entourages $U_n$ of the left uniformity are symmetric entourages such that $U_{n+1} \subset U_n$. The method used in the proof of Proposition 2 of § 1, no. 4 allows us to construct, from the sequence of entourages $(U_n)$, a metric $d$ on $G$ compatible with the left uniformity of $G$; and since for each $z \in G$ the mapping $(x, y) \to (zx, zy)$ leaves each of the $U_n$ invariant, the definition of $d$ shows that it is a left-invariant metric. This method also holds for the right uniformity.

Note that, if the two uniformities of $G$ are distinct, the metric $d$ is not right-invariant, and hence in general $d(x^{-1}, y^{-1}) \neq d(x, y)$.

In particular, if $G$ is a metrizable abelian group, its uniformity is defined by an invariant metric $d$: if $G$ is written additively, we have

We shall often write $|x|$ (or $||x|$) for $d(0, x)$; we have then

The function $|x|$ satisfies the following three conditions:

a) $|-x| = |x|$ for all $x \in G$.

b) $|x + y| \leq |x| + |y|$ for all $x \in G$ and all $y \in G$.

c) $|x| = 0$ if and only if $x = 0$.

Conversely:

The function $d(x, y)$ is a metric on $G$, for the relation $d(x, y) = 0$ is equivalent to $x = y$ by c); we have $d(x, y) = d(y, x)$ by a); and

by b). Moreover, $d$ is invariant, since $(x + z) - (y + z) = x - y$. For each real number $x > 0$, let $V_x$ be the set of all $x \in G$ such that $|x| < x$; then the $V_x$ form a fundamental system $\mathfrak{S}$ of neighbourhoods of $0$ for the topology $\mathscr{T}$, and since $d$ is invariant, $a + \mathfrak{S}$ is a fundamental system of neighbourhoods of $a$ for the topology $\mathscr{T}$, for each $a \in G$. By a), the $V_x$ are symmetric, and by b), we have $V_a + V_x \subset V_{2a}$; hence the topology $\mathscr{T}$ is compatible with the group structure of $G$ (Chapter III, § 1, no. 2). The last part of the proposition follows immediately.

Conditions a', b' and c' are equivalent to c) together with the condition

For a' and b' clearly imply b'); conversely, taking $x = 0$ in b') and using c), we see that $|-y| \leq |y|$; replacing $y$ by $-y$ it follows that $|-y| = |y|$, which is a); replacing $y$ by $-y$ in b') we then get b').

The first part of the proposition is a consequence of the fact that the identity element of G/H has a countable fundamental system of neighbourhoods in G/H; for if (V_n) is a fundamental system of neighbourhoods of e in G, then the canonical images $\dot{V}_n$ of the sets V_n in G/H form a fundamental system of neighbourhoods of the identity element of G/H (Chapter III, § 2, no. 6, Proposition 17).

To show that G/H is complete if G is complete, it is enough, by § 2, no. 6, Proposition 9, to show that every Cauchy sequence $(\dot{x}_n)$ (with respect to the left uniformity of G/H) is convergent. We may assume, by passing to a subsequence of $(\dot{x}_n)$ if necessary, that for each pair of indices $p, q$ such that $p \geq n$ and $q \geq n$, we have $\dot{x}_p^{-1}\dot{x}_q \in \dot{V}_n$; this means that for each pair of points $y \in \dot{x}_p, z \in \dot{x}_q$, we have $y^{-1}z \in HV_n = V_nH$; and therefore, for each $y \in \dot{x}_p$, the intersection of $\dot{x}_q$ and the neighbourhood $yV_n$ of $y$ is not empty. Suppose then that the sequence (V_n) has been chosen so that $V_{n+1}^2 \subset V_n$, and define inductively a sequence $(x_n)$ of points of G, such that $x_n \in \dot{x}_n$ and $x_{n+1} \in x_n V_n$; this is possible by what has been said. It follows then by induction that for each $p > 0$ we have $x_{n+p} \in x_n V_n V_{n+1} \ldots V_{n+p-1} \subset x_n V_{n-1}$. The sequence $(x_n)$ is therefore a Cauchy sequence in G, so it converges to a point $a$; and it follows immediately that the canonical image $\dot{a}$ of $a$ in G/H is the limit of the sequence $(\dot{x}_n)$.

H is a normal subgroup of G (Chapter III, § 2, no. 3, Proposition 8) and Proposition 4 shows that G/H is complete. Also if $\varphi$ is the canonical mapping of G onto G/H, it is clear that $\varphi(G_0)$ is dense in G/H. The result therefore follows from Chapter III, § 2, no. 7, Proposition 21.

Let G, G' be two Hausdorff abelian topological groups, and $\hat{G}, \hat{G}'$ their respective completions. We recall (Chapter III, § 3, no. 3, Proposition 5) that if $u$ is a continuous homomorphism of G into G', then $u$ is uniformly continuous and extends uniquely to a continuous homomorphism of $\hat{G}$ into $\hat{G}'$, which we shall denote by $\hat{u}$ in the remainder

(*) There exist non-metrizable complete groups G containing a closed subgroup H such that G/H is not complete.

of this subsection. The diagram

in which $i, i'$ are the canonical injections) is commutative. If $v$ is a continuous homomorphism of $G'$ into a Hausdorff abelian topological group $G''$, and if $w = v \circ u$, then it is follows immediately that $\hat{w} = \hat{v} \circ \hat{u}$.

Let $H$ be a closed subgroup of $G$ and let $E = G/H$. Let $j : H \to G$ and $p : G \to E$ be the canonical mappings. Let $\overline{H}$ be the closure of $H$ in $\hat{G}$; $\overline{H}$ is a complete group and we identify it with the completion $\hat{H}$ of $H$. The continuous extension $\hat{j}$ of $j$ to $\hat{H}$ is evidently the canonical injection of $\hat{H}$ in $\hat{G}$.

Suppose from now on that $G$ is metrizable. Then the canonical mapping of $E = GH$ into $\hat{G}\hat{H}$ is a topological isomorphism of $E$ onto a dense subgroup of the complete group $\hat{G}/\hat{H}$ (Corollary 1), and thus we may identify $\hat{G}\hat{H}$ with $\hat{E}$ and the continuous extension $\hat{p}$ of $p$ to $\hat{G}$ with the canonical mapping of $\hat{G}$ onto $\hat{G}/\hat{H}$.

Let $u = j \circ v \circ p$ be the canonical factorization of $u$, where $v$ is an isomorphism of the topological group $G/N$ onto the topological group $u(G) = P$. We have $\hat{u} = \hat{j} \circ \hat{v} \circ \hat{p}$, and we have seen that $\hat{p}$ is the canonical mapping of $\hat{G}$ onto $\hat{G}\hat{N}$, and that $\hat{j}$ is the canonical mapping of $\hat{P}$ into $\hat{G}$. On the other hand, $\hat{v}$ is an isomorphism of $\hat{G}/\hat{N}$ onto $\hat{P}$ (Chapter III, § 3, no. 4, Proposition 5), whence the result.

For if we put $N = u(G) = \overline{v}(0)$, it follows from Corollary 2 that $\hat{N}$ is both the image of $\hat{u}$ and the kernel of $\hat{v}$.

The condition is necessary, for the set of all $x \in K$ such that $|x|_1 < 1$ is the same as the set of all $x$ such that, with respect to the topology defined by the absolute value $|x|_1$, $\lim_{n \to \infty} x^n = 0$.

Suppose conversely that $|x|_1 < 1 \Longrightarrow |x|_2 < 1$. Then $|x|_1 > 1 \Longrightarrow |x|_2 > 1$, because $|x^{-1}|_1 < 1$ and therefore $|x^{-1}|_2 < 1$. Since by hypothesis the absolute value $|x|_1$ is not improper, there exists $x_0 \in K$ such that $|x_0|_1 > 1$. Let $a = |x_0|_1, b = |x_0|_2$ and let $\rho = \log b / \log a > 0$. Let $x \in K^*$ and put $|x|_1 = |x_0|_1^\gamma$. If $m$ and $n$ are integers such that $n > 0$ and $m/n > \gamma$, then $|x|_1 < |x_0|_1^{m/n}$, and therefore $|x^n x_0^{-m}|_1 < 1$; hence $|x^n x_0^{-m}|_2 < 1, |x|_2 < |x_0|_2^{m/n}$. Similarly, if $m/n < \gamma$, we see that $|x|_2 > |x_0|_2^{m/n}$; it follows therefore that $|x|_2 = |x_0|_2^\gamma$; in other words

i.e., $|x|_2 = |x|^p_1$. It is now clear that the neighbourhoods of zero for the topologies defined on $K$ by $|x|_1$ and $|x|_2$ are identical.

Conversely, if $|x|$ is any absolute value on $K$, the function $|x|^p$ is an absolute value on $K$ (equivalent to $|x|$) for all $p$ such that $0 < p \leq 1$. We have only to verify the inequality $|x + y|^p \leq |x|^p + |y|^p$; and since $|x + y|^p \leq (|x| + |y|)^p$ it is enough to show that, if $a > 0$ and $b > 0$, we have $(a + b)^p \leq a^p + b^p$ for any $p$ such that $0 < p \leq 1$. If we put $c = a/(a + b)$ and $d = b/(a + b)$, we have $c + d = 1$, and the inequality to be proved is $c^p + d^p \geq 1$; but this follows immediately from the relations $c^p \geq c$ and $d^p \geq d$, which are valid since $0 < c \leq 1$, $0 < d \leq 1$, $0 < p \leq 1$.

Hence the set of values of $r > 0$ such that $|x|^r$ is an absolute value is a finite or infinite interval of $\mathbf{R}$ with left-hand end-point $0$; if it is finite, it is evidently closed; for if we have $|x + y|^r \leq |x|^r + |y|^r$ for any $x, y$ in $K$ and all $r$ such that $0 < r < r_0$, then by continuity the inequality is still valid for $r = r_0$. If $|x|^r$ is an absolute value for all $r > 0$, then we have

for all $x$ and $y$ in $K$ and all $r > 0$. Now, if $a, b$ are two real numbers $\geq 0$, we have $\lim_{r \to \infty} (a^r + b^r)^{1/c} = \sup (a, b)$; for, supposing for example that $a \geq b$, we have $a \leq (a^r + b^r)^{1/r} \leq 2^{1/r} a$, and the result follows by letting $r \to + \infty$.

Thus, if $|x|^r$ is an absolute value for all $r > 0$, we have

which can be expressed by saying that $v(x) = -\log |x| \ (x \neq 0)$ is a valuation on $K$.

Let $\mathfrak{F}$ be a Cauchy filter on $K$ (with respect to the additive uniformity) which does not have $0$ as a cluster point; to show that $\hat{K}$ is a division ring it is enough to establish that the image of $\mathfrak{F}$ under the mapping $x \to x^{-1}$ is a Cauchy filter base (Chapter III, § 6, no. 8, Proposition 7). Now, by hypothesis there exists a real number $\alpha > 0$ and a set $A \in \mathfrak{F}$ such that $|x| \geq \alpha$ for all $x \in A$; on the other hand, for each $\varepsilon > 0$ there exists a set $B \in \mathfrak{F}$ such that $B \subset A$ and $|x - y| \leq \varepsilon$ for all $x \in B$.

and $y \in B$; hence

and the first part of the proposition follows. The invariant metric $|x - y| = d(x, y)$ extends by continuity to a metric on $\hat{K}$ (§ 2, no. 1, Proposition 1) which defines the topology of $\hat{K}$ and is invariant by the principle of extension of identities; we continue to denote this invariant metric by $d(x, y)$. If we put $|x| = d(0, x)$ for $x \in \hat{K}$, it is clear that $|x|$ is the extension by continuity of the function $|x|$ on $K$ and is therefore an absolute value on $\hat{K}$ by the principle of extension of identities.

(I)

for all $x \in E$.

These inequalities are sufficient, for it follows from the relation $a . p(x) \leq q(x)$ that, for each $r > 0$, the closed ball with centre o and radius $ar$ (relative to the norm $q$) is contained in the closed ball with centre o and radius $r$ (relative to the norm $p$); hence the topology defined by $q$ is finer than the topology defined by $p$. Similarly the inequality $q(x) \leq b . p(x)$ shows that the topology defined by $p$ is finer than that defined by $q$, and hence $p$ and $q$ are equivalent.

Let us now show that the inequalities (I) are necessary. If the topology defined by $q$ is finer than the topology defined by $p$, then the unit ball with respect to $p$ contains a closed ball with centre o and radius $\alpha > 0$ with respect to $q$; i.e., the relation $q(x) \leq \alpha$ implies $p(x) \leq 1$. If $t_0 \in K$ is such that $0 < |t_0| < 1$, then for each $x \neq o$ in $E$ there is a unique rational integer $k$ such that $\alpha |t_0| < q(t_0^k x) \leq \alpha$; therefore $p(t_0^k x) \leq 1$, so that

putting $a = \alpha |t_0|$ we have therefore $a . p(x) \leq q(x)$ for all $x \neq o$, and this inequality is also valid when $x = o$. Similarly we show that if the topology defined by $p$ is finer than that defined by $q$, there exists $b > 0$ such that $q(x) \leq b . p(x)$.

The extension of $tx$ by continuity is a particular case of the theorem of extension of a continuous bilinear mapping of a product of two abelian groups into a third (Chapter III, § 6, no. 5, Theorem 1); we have $1.x = x$ and $t(u x) = (t u)x$ for $t \in \hat{K}$, $u \in \hat{K}$ and $x \in \hat{E}$, by the principle of extension of identities; hence the external law $(t, x) \to t x$ indeed defines on $\hat{E}$ a structure of a vector space over $\hat{K}$. On the other hand, the invariant metric $\overline{d}(x, y) = \overline{p}(x - y)$ extends to an invariant metric $\overline{d}$ on $\hat{E}$ (§ 2, no. 1, Proposition 1) which defines the topology of $\hat{E}$; if we set $\overline{p}(x) = \overline{d}(0, x)$, then $\overline{p}$ is the extension of $p$ by continuity, and satisfies axioms (NO_I) and (NO_{II}); by virtue of the continuity of $t x$ on $\hat{K} \times \hat{E}$, $\overline{p}$ also satisfies (NO_{III}) (principle of extension of identities) and is therefore a norm on $\hat{E}$.

When we have to consider a definite normed space structure on a vector space $E$, we shall usually denote the norm of a vector $x$ by $||x||$, unless this notation is likely to lead to confusion.