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Let $\mathcal{A}$ be a complex associative Hausdorff topological algebra, and let $A\subset\mathcal{A}$ be a locally compact Abelian (LCA) subgroup (multiplicative). The linear span $\mathbb{C}\{A\}$ has an induced topology as a subalgebra of $\mathcal{A}$.

On the other hand, let $\hat A$ be the Pontrjiagin dual, so that $\phi:A\mapsto\hat{\hat A}\subset C(\hat A)$ is an isomorphism of LCA groups, where $C(\hat A)$ is given the topology of compact convergence. Now $\mathbb{C}\{A\}$ can be given the compact convergence topology of the subalgebra $\mathbb{C}\{\phi(A)\}\subset C(\hat A)$.

Question: Are the two topologies of the algebra $\mathbb{C}\{A\}$ above equivalent? More generally, how unique is the topology of $\mathbb{C}\{A\}$ given the LCA group $A$?

I assume that elements of $A$ are linearly independent in $\mathcal{A}$.

Thank you.

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  • $\begingroup$ What do you mean by "elements of $A$ are linearly independent in $\mathcal{A}$"? it sounds absurd. $\endgroup$
    – YCor
    May 29, 2017 at 11:15
  • $\begingroup$ What exactly does sound to you like absurd? $\endgroup$
    – Bedovlat
    May 29, 2017 at 12:09
  • $\begingroup$ If you have a linear space $\mathcal{A}$ and a subset $Z$, to say that "element of $Z$ are linearly independent in $\mathcal{A}$ has some standard sense", namely there is no nontrivial linear combination between them. If $0\in Z$, or if $Z$ contains some vector $v$ as well as $2v$, this is not the case. $\endgroup$
    – YCor
    May 29, 2017 at 12:22
  • $\begingroup$ First of all, $A\subset\mathcal{A}$ is a subGROUP, so $0\in A$ is excluded. Of course, it may happen that $A$ is not a linearly independent set. That is exactly why I have written that sentence. So, again, what is abusrd here? $\endgroup$
    – Bedovlat
    May 29, 2017 at 13:01
  • $\begingroup$ Ok, you might have confused an additive subgroup with a multiplicative subgroup. If that is the case, I can understand. I will amend the statement. $\endgroup$
    – Bedovlat
    May 29, 2017 at 13:07

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