# $S^{3}$-valued harmonic analysis

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

• This isn't enough for a full answer, but it seems straightforward that for $\Gamma=\mathbb Z$, the space $\tilde{\tilde{\Gamma}}$ is just $S^3$ again, by the same proof as in the classical case of $S^1$ (since $\mathbb Z$ is cyclic and generated by $1$, the "character" $\phi$ determines (and is determined by) the value $\phi(1)\in S^3$, so we get a bijection, which is continuous by the definition of the topology, and hence a homeomorphism by virtue of compactness.) – Mike Jury Aug 26 '14 at 8:09
• @MikeJury yes thank you. About $\tilde{\tilde{S^{1}}}$ one can find, at least 3 copy of $\mathbb{Z}$ in it. But how they match to each other? – Ali Taghavi Aug 26 '14 at 8:26
• Isn't it more complicated than that? If we choose real numbers $b,c,d$ satisfying $b^2+c^2+d^2=1$ and form the quaternion $q=bi+cj+dk$, then $q^2=-1$, and $\phi(x+iy)=x+yq$ is homomorphism of $S^1$ into $S^3$? – Mike Jury Aug 26 '14 at 8:38
• How is homomorphisms into $S^3$ a group? I only see a topological space. You need commutativity of $S^1$ for the Pontryagin dual group structure. – nsrt Aug 26 '14 at 15:07
• Continuous group homomorphisms from $S^1$ to a Lie group are automatically smooth (this is related to Hilbert's fifth problem, and I think was proved by Brouwer), so indeed @MikeJury wrote down all injective group homomorphisms $S^1\to S^3$, and this space is homeomorphic to $S^2$. The space of all group homomorphisms is now obtained by composing with all group homomorphisms $S^1\to S^1$. So the space we obtain is the union of a point and a countable union of copies of $S^2$. – nsrt Aug 27 '14 at 8:35