4
$\begingroup$

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

$\endgroup$
  • 2
    $\begingroup$ 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.) $\endgroup$ – Mike Jury Aug 26 '14 at 8:09
  • $\begingroup$ @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? $\endgroup$ – Ali Taghavi Aug 26 '14 at 8:26
  • 1
    $\begingroup$ 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 $S^3$? $\endgroup$ – Mike Jury Aug 26 '14 at 8:38
  • 3
    $\begingroup$ 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. $\endgroup$ – nsrt Aug 26 '14 at 15:07
  • 1
    $\begingroup$ 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 $S^2$. $\endgroup$ – nsrt Aug 27 '14 at 8:35

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Browse other questions tagged or ask your own question.