# Completion of abelian topological groups

During some exercises I came upon the following questions that I cannot answer myself. Given a topological group $G$ we can build the completion using cauchy sequences and denote this completion by $\bar{G}$. This is again a topological group. Let $\phi : G \rightarrow \bar{G}$ be the usual map sending each element of $G$ to the equivalence class of the constant sequence. I know (not hard to show) that $\phi$ is injective if and only if $G$ is hausdorff. My questions are as follows: (EDIT: Assuming $G$ is abelian and satisfies the first axiom of countability)

If $G$ is hausdorff, is $\bar{G}$ hausdorff? Or is $\bar{G}$ automatically hausdorff? (Elements that cannot be seperated in $G$ are mapped to the same equivalence class)

When is $\bar{G}$ isomorphic to $\bar{\bar{G}}$ and how can I show that?

I'd also appreciate it, if someone could point me to some good literature.

Best regards

• You probably want your group to be first-countable, otherwise the completion is not defined in terms of Cauchy sequences. For a very general treatment I suggest Bourbaki, General Topology, Chap. III, §3. – abx Dec 3 '13 at 14:59
• I'm afraid there's an issue: there is a left-completion and a right completion (the left is completing with respect to nets s.t. $g_i^{-1}g_j\to 1$); in case they do not coincide the left completion is not a topological group. – YCor Dec 3 '13 at 16:01
• @Yves Doesn't that coincide in the abelian case? (I forgot to mention abelian above) – Horstenson Dec 3 '13 at 16:42
• @abx Thanks for the heads up, I'll edit and look it up – Horstenson Dec 3 '13 at 16:43
• @Hortenson: of course yes, in the abelian case, but this is a huge restriction. – YCor Dec 3 '13 at 17:49