Remark 1

If G is a separated topological group, $\iota (G)$ is closed in $G^*$.

Let indeed $h\in G^*$ be such that $h\notin \iota (G)$, let $(t_i)_{0\leqslant i\leqslant n}$ be a sequence adapted to $h$, put $\varepsilon =$ min$_{1\leqslant i\leqslant n}(t_i-t_{i-1})$. Let V be a neighbourhood of $e$ in G such that $h(t_i)^{-1}h(t_j)\notin V$, for every pair $(i, j)$ of integers such that $0\leqslant i, j\leqslant n-1$ and $h(t_i)=\not h(t_j)$; there exists one since G is separated. Let W be a neighbourhood of $e$ in G such that $W\cdot W^{-1}\subset V$. Let us prove that then $hW^*_{\varepsilon}$ does not meet $\iota (G)$.

Let us reason by contradiction. Let $f$ be an element of $W^*_{\varepsilon}$ and $g$ an element of G such that $hf$ = $\iota (g)$. We therefore have $f(t) =h(t)^{-1}g$ for all $t\in [0,1[$, so that the sequence $(t_i)$ is also adapted to the function $f$. If the value taken by $f$ on the interval $[t_{i-1}, t_i[$ does not belong to W, then $t_i-t_{i-1}< \varepsilon$, since $f\in W^*_{\varepsilon}$. This inequality being false, by definition of $\varepsilon$, we have $f(t)\in W$ for all $t\in [0,1[$. Let $i$ and $j$ then be elements of $\{0, . . . , n-1\}$ such that $h(t_i)=\not h(t_j)$; we have

which contradicts the choice of W. The subgroup $\iota (G)$ of $G^*$ is therefore closed.

Suppose that G is a metrisable topological group and let $d$ be a distance on G which defines its topology. Let $h$ and $h'\in G^*$ and let $(t_i)_{0\leqslant i\leqslant n}$ be a sequence adapted to $h$ and $h'$. The real number

does not depend on the sequence $(t_i)$ chosen; let us denote it by $d^*(h, h')$.

Let us prove that $d^*$ is a distance on $G^*$. We have $d^*(h, h') =$ $d^*(h', h)$ for $h,h'\in G^*$, and $d^*(h, h) = 0$ for all $h\in G^*$. Conversely, let $h$ and $h'$ be elements of $G^*$ such that $d^*(h, h') = 0$. Let $(t_i)_{0\leqslant i\leqslant n}$ be a sequence adapted to $h$ and $h'$. Since

and all the terms of this sum are positive or zero, we have $d(h(t_{i-1}), h'(t_{i-1})) = 0$ for all $i\in \{1, . . . , n\}$, whence $h=h'$. Finally, let $h,h',h''$ be elements of $G^*$ and let $(t_i)_{0\leqslant i\leqslant n}$ be a sequence adapted to each of them. Then,

therefore $d^*$ satisfies the triangle inequality.

This distance $d^*$ is invariant under right (resp. left) translations if $d$ is.