2
$\begingroup$

Let $G$ be a topological abelian group (linearly topologized), not necessarilly locally compact and define its dual:

$$\widehat{G}:=\operatorname{Hom}(G,S^1)$$

We endow $\widehat{G}$ with the compact-open topology. Moreover For any subset $S\subset G$, let's introduce the following notation: $$S^\circ:=\{f\in \widehat{G}: f(S)=1\}\subset \widehat{G}\,.$$


My question is the following: Consider the family of subgroups of $\widehat{G}$:

$$\mathcal F:=\{C^\circ:C\subset G \text{ is closed and compact}\}$$

Is it a local basis at $1$ of closed subgroups? In any case, what is the relationship between closed subsets (or subgroups) of $\widehat{G}$ and $\mathcal F$?

$\endgroup$
1
$\begingroup$

This answer is partial. I hope it will be useful for others.

I understood that $G$ is linearly topologized as there exists a local base $\mathcal B$ at $1$ consisting of open subgroups of $G$. That is $\bigcap\mathcal B=\{1\}$ and for each $H,H’\in\mathcal B$, $H\cap H’\in\mathcal B$. Next, by $\operatorname{Hom}(G, S^1)$ I understood a group of continuous homomorphisms from $G$ to the unit circle $S^1=\{|z|\in\Bbb C:|z|=1\}$ endowed with the topology and multiplication inherited from usual these of $\Bbb C$.

Put $V_0=\{z\in S^1:\operatorname{Re} z>0\}$. If $f\in \widehat{G}$ then there exist $H\in\mathcal B$ such that $f(H)\subset V_0$. Since $V_0$ has no non-trivial subgroups, we have $f(H)=\{1\}$. Conversely, any homomorphism $f:G\to S^1$ such that $f(H)=\{1\}$ for some $H\in\mathcal B$, is continuous. Thus for any $f\in\widehat G$ we have $\operatorname{ker} f\in\mathcal B$.

Then for any nonempty subset $S$ of $G$, $S^\circ=\{f\in \widehat{G}: S\subset \operatorname{ker} f\}=\{f\in \widehat{G}: \langle S\rangle\subset \operatorname{ker} f\}=\langle S\rangle^\circ,$ where $\langle S\rangle$ is the subgroup of $G$ generated by $S$. Thus $\mathcal F=\{H^\circ: H$ is a compactly generated subgroup of $G\}$.

I assume that the operation of $\widehat G$ is defined by putting $(fg)(x)=f(x)g(x)$ for each $f,g\in\widehat G$ and $x\in G$. Then the identity of $\widehat G$ is the annulating homomorphism which maps each element of $G$ to $1$.

Thus $S^\circ$ is a subgroup of $G$ for any nonempty subset $S$ of $G$.

Is $\mathcal F$ a local basis at $1$ of closed subgroups?

The local base $\widehat{\mathcal B}$ at $1$ of $\widehat G$ consists of the sets $(C,V)$, where $C$ is a compact subset of $G$ and $V\subset S^1$ is an open neighborhood of $1$. Clearly, for any $(C,V)\in \widehat{\mathcal B}$ we have $C^\circ\subset (C,V)$. I don’t know whether for any non-empty compact subset $C$ of $G$ there exists a compact subset $C’$ of $G$ and an open neighborhood $V\subset S^1$ of $1$ such that $(C’,V)\subset C^\circ$.

what is the relationship between closed subsets (or subgroups) of $\widehat{G}$ and $\mathcal F$?

Clearly, $S^\circ$ is a subgroup of $\widehat{G}$ for any non-empty subset $S$ of $G$. Moreover, $S^\circ$ is closed. Indeed, suppose to the contrary that there exists $f\in \overline{S^\circ}$ and $x\in S$ such that $f(x)\ne 1$. Then $(\{x\},S^1\setminus\{1\})$ is an open neighborhood of $f$, so there exists an element $g\in (\{x\},S^1\setminus\{1\})\cap S^\circ$. Since $g\in (\{x\},S^1\setminus\{1\})$, $g(x)\ne 1$. On the other hand, since $g\in S^\circ $ and $x\in S$, $g(x)=1$, a contradiction. I don’t know whether each closed subgroup of $\widehat G$ belongs to $\mathcal F$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.