**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).

- 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 $C(\tilde{\tilde{\Gamma}})$?