I studied a completion of topological group by textbook Atiyah. :
What I just understand is
if $G$ is a topological group, then if $\hat{G}$ is a set of equivalent Cauchy sequence, then we call $\hat{G}$ as a completion of $G$. I understand and verify that $\hat{G}$ has a topological group structure. After that, if $G$ is any "group" (not necessarily a topological space), then we can consider the sequence of subgroups
$$ G=G_0 \supseteq G_1 \supseteq\cdots G_n \supseteq \cdots $$
What I understand is that we "can give a topology on G", by taking $U$ is nbd of $0$ iff $U$ contains some $G_n$ (then translate it by all points in $G$). After give a topology on $G$ in this way, I also understand that
$$ \hat{G} \cong \varprojlim G/G_n$$, and what is explicit isomorphism.
Here is my question.
- Am I correctly understand that we can give a topology on arbitrary group in this way?
- If $G$ is already a topological group, then there exists appropriate sequence of subgroups which induces a topology same with original one ?
Any helps would be greatly appreciated. Thank you!
Sure. But that depends on the choice of the subgroup chain. Arbitrary chain may not be interesting. For example, for any group $G$ we have the following sequence $G\supseteq E\supseteq E\supseteq\cdots$ where $E\leq G$ denotes the trivial subgroup. The resulting topology is discrete though.
No. Consider $\mathbb{R}$ with the Euclidean topology. Note that the only subgroup of $\mathbb{R}$ inside $(-1,1)$, and hence any small enough neighbourhood of $0$, is the trivial subgroup $E=\{0\}$. Therefore the only way to give such a chain of subgroups is with the trivial chain $\mathbb{R}\supseteq E\supseteq E\supseteq\cdots$. This however generates the discrete topology instead of Euclidean.