Timeline for $S^{3}$-valued harmonic analysis
Current License: CC BY-SA 3.0
11 events
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| Jun 15, 2020 at 7:27 | history | edited | CommunityBot |
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| Aug 27, 2014 at 8:35 | comment | added | nsrt | 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 $S^2$. | |
| Aug 26, 2014 at 16:50 | history | edited | Ali Taghavi | CC BY-SA 3.0 |
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| Aug 26, 2014 at 16:45 | comment | added | Ali Taghavi | @nsrt Yes. Thank you for your comment. I was incorrect. The pointwise multiplication does not give us a group structure since $S^{3}$ is not abelian. I revise the question | |
| Aug 26, 2014 at 15:07 | comment | added | nsrt | 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. | |
| Aug 26, 2014 at 8:38 | comment | added | Mike Jury | 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 $S^3$? | |
| Aug 26, 2014 at 8:26 | comment | added | Ali Taghavi | @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? | |
| Aug 26, 2014 at 8:09 | comment | added | Mike Jury | 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.) | |
| Aug 26, 2014 at 7:53 | history | edited | Ali Taghavi | CC BY-SA 3.0 |
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| Aug 26, 2014 at 7:12 | history | edited | Ali Taghavi | CC BY-SA 3.0 |
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| Aug 26, 2014 at 7:03 | history | asked | Ali Taghavi | CC BY-SA 3.0 |