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Edit:

Note that $S^{3}$ with the quaternion operation is a group. For a locally compact Abelian group $\Gamma$ we consider $$\tilde{\tilde{\Gamma}}=\{\phi:\Gamma \to S^{3} \mid \phi(xy)=\phi(x)\phi(y) \}$$ This is the $S^{3}$ valued analogue of $$\tilde{\Gamma}=\{\phi:\Gamma\to S^{1}\mid\phi(xy)=\phi(x)\phi(y)\}$$

As nsrt pointed out in his comment this is not more than an space(not necessarily a group).We equip space with compact open topology. It seems that one can prove that $\tilde{\tilde{\Gamma}}$ is compact(discrete) if $\Gamma$ is discrete(compact).

  1. Does $\tilde{\tilde {\Gamma}}$ have a known topological structure for $\Gamma=\mathbb{Z}$ or $\Gamma=S^{1}$. (Is it homeomorphic to a known topological space?)
 

2.Motivated by Pontryagin theorem for $S^{1}$ valued harmonic analysis, we ask: Is it true that This topological space is connected $\iff$ $\Gamma$ is torsion free.

 

3.We know that $C_{red}^{*} \Gamma$ is an operator theoretical version of $C(\tilde{\Gamma})$. Now we ask: What is a non commutative( and operator theoretical) analogues of $C(\tilde{\tilde{\Gamma}})$?

Edit:

Note that $S^{3}$ with the quaternion operation is a group. For a locally compact Abelian group $\Gamma$ we consider $$\tilde{\tilde{\Gamma}}=\{\phi:\Gamma \to S^{3} \mid \phi(xy)=\phi(x)\phi(y) \}$$ This is the $S^{3}$ valued analogue of $$\tilde{\Gamma}=\{\phi:\Gamma\to S^{1}\mid\phi(xy)=\phi(x)\phi(y)\}$$

As nsrt pointed out in his comment this is not more than an space(not necessarily a group).We equip space with compact open topology. It seems that one can prove that $\tilde{\tilde{\Gamma}}$ is compact(discrete) if $\Gamma$ is discrete(compact).

  1. Does $\tilde{\tilde {\Gamma}}$ have a known topological structure for $\Gamma=\mathbb{Z}$ or $\Gamma=S^{1}$. (Is it homeomorphic to a known topological space?)
 

2.Motivated by Pontryagin theorem for $S^{1}$ valued harmonic analysis, we ask: Is it true that This topological space is connected $\iff$ $\Gamma$ is torsion free.

 

3.We know that $C_{red}^{*} \Gamma$ is an operator theoretical version of $C(\tilde{\Gamma})$. Now we ask: What is a non commutative( and operator theoretical) analogues of $C(\tilde{\tilde{\Gamma}})$?

Edit:

Note that $S^{3}$ with the quaternion operation is a group. For a locally compact Abelian group $\Gamma$ we consider $$\tilde{\tilde{\Gamma}}=\{\phi:\Gamma \to S^{3} \mid \phi(xy)=\phi(x)\phi(y) \}$$ This is the $S^{3}$ valued analogue of $$\tilde{\Gamma}=\{\phi:\Gamma\to S^{1}\mid\phi(xy)=\phi(x)\phi(y)\}$$

As nsrt pointed out in his comment this is not more than an space(not necessarily a group).We equip space with compact open topology. It seems that one can prove that $\tilde{\tilde{\Gamma}}$ is compact(discrete) if $\Gamma$ is discrete(compact).

  1. Does $\tilde{\tilde {\Gamma}}$ have a known topological structure for $\Gamma=\mathbb{Z}$ or $\Gamma=S^{1}$. (Is it homeomorphic to a known topological space?)

2.Motivated by Pontryagin theorem for $S^{1}$ valued harmonic analysis, we ask: Is it true that This topological space is connected $\iff$ $\Gamma$ is torsion free.

3.We know that $C_{red}^{*} \Gamma$ is an operator theoretical version of $C(\tilde{\Gamma})$. Now we ask: What is a non commutative( and operator theoretical) analogues of $C(\tilde{\tilde{\Gamma}})$?

added 106 characters in body
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Ali Taghavi
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Edit:

Note that $S^{3}$ with the quaternion operation is a group. For a locally compact Abelian group $\Gamma$ we consider $$\tilde{\tilde{\Gamma}}=\{\phi:\Gamma \to S^{3} \mid \phi(xy)=\phi(x)\phi(y) \}$$ This is the $S^{3}$ valued analogue of $$\tilde{\Gamma}=\{\phi:\Gamma\to S^{1}\mid\phi(xy)=\phi(x)\phi(y)\}$$

We equip this non abelianAs nsrt pointed out in his comment this is not more than an space(not necessarily a group).We equip space with compact open topology. ObviouslyIt seems that one can prove that $\tilde{\tilde{\Gamma}}$ is compact(discrete) if $\Gamma$ is discrete(compact).

  1. Does $\tilde{\tilde {\Gamma}}$ have a known topological structure for $\Gamma=\mathbb{Z}$ or $\Gamma=S^{1}$. (Is it homeomorphic to a known topological space?)

2.Motivated by Pontryagin theorem for $S^{1}$ valued harmonic analysis, we ask: Is it true that This group topological space is connected $\iff$ $\Gamma$ is torsion free.

3.We know that $C_{red}^{*} \Gamma$ is an operator theoretical version of $C(\tilde{\Gamma})$. Now we ask: What is a non commutative( and operator theoretical) analogues of $C(\tilde{\tilde{\Gamma}})$?

Note that $S^{3}$ with the quaternion operation is a group. For a locally compact Abelian group $\Gamma$ we consider $$\tilde{\tilde{\Gamma}}=\{\phi:\Gamma \to S^{3} \mid \phi(xy)=\phi(x)\phi(y) \}$$ This is the $S^{3}$ valued analogue of $$\tilde{\Gamma}=\{\phi:\Gamma\to S^{1}\mid\phi(xy)=\phi(x)\phi(y)\}$$

We equip this non abelian group with compact open topology. Obviously $\tilde{\tilde{\Gamma}}$ is compact(discrete) if $\Gamma$ is discrete(compact).

  1. Does $\tilde{\tilde {\Gamma}}$ have a known topological structure for $\Gamma=\mathbb{Z}$ or $\Gamma=S^{1}$. (Is it homeomorphic to a known topological space?)

2.Motivated by Pontryagin theorem for $S^{1}$ valued harmonic analysis, we ask: Is it true that This group is connected $\iff$ $\Gamma$ is torsion free.

3.We know that $C_{red}^{*} \Gamma$ is an operator theoretical version of $C(\tilde{\Gamma})$. Now we ask: What is a non commutative( and operator theoretical) analogues of $C(\tilde{\tilde{\Gamma}})$?

Edit:

Note that $S^{3}$ with the quaternion operation is a group. For a locally compact Abelian group $\Gamma$ we consider $$\tilde{\tilde{\Gamma}}=\{\phi:\Gamma \to S^{3} \mid \phi(xy)=\phi(x)\phi(y) \}$$ This is the $S^{3}$ valued analogue of $$\tilde{\Gamma}=\{\phi:\Gamma\to S^{1}\mid\phi(xy)=\phi(x)\phi(y)\}$$

As nsrt pointed out in his comment this is not more than an space(not necessarily a group).We equip space with compact open topology. It seems that one can prove that $\tilde{\tilde{\Gamma}}$ is compact(discrete) if $\Gamma$ is discrete(compact).

  1. Does $\tilde{\tilde {\Gamma}}$ have a known topological structure for $\Gamma=\mathbb{Z}$ or $\Gamma=S^{1}$. (Is it homeomorphic to a known topological space?)

2.Motivated by Pontryagin theorem for $S^{1}$ valued harmonic analysis, we ask: Is it true that This topological space is connected $\iff$ $\Gamma$ is torsion free.

3.We know that $C_{red}^{*} \Gamma$ is an operator theoretical version of $C(\tilde{\Gamma})$. Now we ask: What is a non commutative( and operator theoretical) analogues of $C(\tilde{\tilde{\Gamma}})$?

added 1 character in body
Source Link
Ali Taghavi
  • 356
  • 8
  • 31
  • 123

Note that $S^{3}$ with the quaternion operation is a group. For a locally compact Abelian group $\Gamma$ we consider $$\tilde{\tilde{\Gamma}}=\{\phi:\Gamma \to S^{3} \mid \phi(xy)=\phi(x)\phi(y) \}$$ This is the $S^{3}$ valued analogue of $$\tilde{\Gamma}=\{\phi:\Gamma\to S^{1}\mid\phi(xy)=\phi(x)\phi(y)\}$$

We equip this non abelian group with compact open topology. Obviously $\tilde{\tilde{\Gamma}}$ is compact(discrete) if $\Gamma$ is discrete(compact).

  1. Does $\tilde{\tilde {\Gamma}}$ have a known topological structure for $\Gamma=\mathbb{Z}$ or $\Gamma=S^{1}$. (Is it homeomorphic to a known topological space?)

2.Motivated by Pontryagin theorem for $S^{1}$ valued harmonic analysis, we ask: Is it true that This group is connected $\iff$ $\Gamma$ is torsion free.

3.We nowknow that $C_{red}^{*} \Gamma$ is an operator theoretical version of $C(\tilde{\Gamma})$. Now we ask: What is a non commutative( and operator theoretical) analogues of $C(\tilde{\tilde{\Gamma}})$?

Note that $S^{3}$ with the quaternion operation is a group. For a locally compact Abelian group $\Gamma$ we consider $$\tilde{\tilde{\Gamma}}=\{\phi:\Gamma \to S^{3} \mid \phi(xy)=\phi(x)\phi(y) \}$$ This is the $S^{3}$ valued analogue of $$\tilde{\Gamma}=\{\phi:\Gamma\to S^{1}\mid\phi(xy)=\phi(x)\phi(y)\}$$

We equip this non abelian group with compact open topology. Obviously $\tilde{\tilde{\Gamma}}$ is compact(discrete) if $\Gamma$ is discrete(compact).

  1. Does $\tilde{\tilde {\Gamma}}$ have a known topological structure for $\Gamma=\mathbb{Z}$ or $\Gamma=S^{1}$. (Is it homeomorphic to a known topological space?)

2.Motivated by Pontryagin theorem for $S^{1}$ valued harmonic analysis, we ask: Is it true that This group is connected $\iff$ $\Gamma$ is torsion free.

3.We now that $C_{red}^{*} \Gamma$ is an operator theoretical version of $C(\tilde{\Gamma})$. Now we ask: What is a non commutative( and operator theoretical) analogues of $C(\tilde{\tilde{\Gamma}})$?

Note that $S^{3}$ with the quaternion operation is a group. For a locally compact Abelian group $\Gamma$ we consider $$\tilde{\tilde{\Gamma}}=\{\phi:\Gamma \to S^{3} \mid \phi(xy)=\phi(x)\phi(y) \}$$ This is the $S^{3}$ valued analogue of $$\tilde{\Gamma}=\{\phi:\Gamma\to S^{1}\mid\phi(xy)=\phi(x)\phi(y)\}$$

We equip this non abelian group with compact open topology. Obviously $\tilde{\tilde{\Gamma}}$ is compact(discrete) if $\Gamma$ is discrete(compact).

  1. Does $\tilde{\tilde {\Gamma}}$ have a known topological structure for $\Gamma=\mathbb{Z}$ or $\Gamma=S^{1}$. (Is it homeomorphic to a known topological space?)

2.Motivated by Pontryagin theorem for $S^{1}$ valued harmonic analysis, we ask: Is it true that This group is connected $\iff$ $\Gamma$ is torsion free.

3.We know that $C_{red}^{*} \Gamma$ is an operator theoretical version of $C(\tilde{\Gamma})$. Now we ask: What is a non commutative( and operator theoretical) analogues of $C(\tilde{\tilde{\Gamma}})$?

added 330 characters in body; edited tags
Source Link
Ali Taghavi
  • 356
  • 8
  • 31
  • 123
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Source Link
Ali Taghavi
  • 356
  • 8
  • 31
  • 123
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