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Bedovlat
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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.

Let $\mathcal{A}$ be a complex associative Hausdorff topological algebra, and let $A\subset\mathcal{A}$ be a locally compact Abelian (LCA) subgroup. 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.

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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Bedovlat
Bedovlat
  • 2k
  • 9
  • 13

Let $\mathcal{A}$ be a complex associative Hausdorff topological algebra, and let $A\subset\mathcal{A}$ be a locally compact Abelian (LCA) subgroup. 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.

Let $\mathcal{A}$ be a complex associative Hausdorff topological algebra, and let $A\subset\mathcal{A}$ be a locally compact Abelian (LCA) subgroup. 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 $A$?

Thank you.

Let $\mathcal{A}$ be a complex associative Hausdorff topological algebra, and let $A\subset\mathcal{A}$ be a locally compact Abelian (LCA) subgroup. 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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Bedovlat
Bedovlat
  • 2k
  • 9
  • 13

Topology of the algebra $\mathbb{C}\{A\}$ for a LCA group $A$

Let $\mathcal{A}$ be a complex associative Hausdorff topological algebra, and let $A\subset\mathcal{A}$ be a locally compact Abelian (LCA) subgroup. 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 $A$?

Thank you.