Intuitive visualization or understanding from $S^2$ to $SO(3)$ and $S^3$ -- and where to find an $S^1$? We know that

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*the rotational operations on the $S^2$ generates a $SO(3)=RP^3=S^3/Z_2=SU(2)/Z_2$.


*The Hopf fibration of $S^1$ over $S^2$ can generate $S^3=SU(2)$.
My questions are that -

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*How to find the $S^1$ (mentioned in the description 2.) from the first description 1? What is the role of $S^1$ in the picture 1. of rotational operations on the $S^2$ generates a $SO(3)$?


*How to find the role of $SO(3)$ in the picture 2. of the Hopf fibration of $S^1$ over $S^2$ as $S^3$?
 A: If a group $G$ acts on a set $\Omega$, and $\omega\in\Omega$ is a point, then we may state the orbit-stabilizer theorem: the orbit $\mathrm{Orb}(\omega)$, as a $G$-set, is equivalent to the coset space $G/\mathrm{Stab}(\omega)$. The correspondence between them relates the coset $g\mathrm{Stab}(\omega)$ to the translate $g\omega$.
We can view $S^3$ as the set of versors, or unit quaternions, and $S^2$ as the set of unit vectors. Then $S^3$ acts on $S^2$ by conjugation. Indeed, conjugating by $\exp(\theta\mathbf{v})$ has the effect of rotating around $\mathbf{v}$ by $2\theta$, so the kernel of this action is the unit reals $S^0=\{\pm1\}$, and the action is a homomorphism $S^3\to\mathrm{SO}(3)$.
As $S^3$ acts on $S^2$, the stabilizer of $\mathbf{i}$ is the set $S^1\subset\mathbb{C}$ of phasors, or unit complex numbers. Therefore, we can say that $S^3/S^1\cong S^2$ are equivalent as $S^3$-sets. Or, equivalently, there is a fiber bundle
$$ S^1 \to S^3 \to S^2. $$
In particular, the map $S^3\to S^2$ sends $p\mapsto p\mathbf{i}p^{-1}$, and the fibers are cosets of $S^1$ within $S^3$. There is artwork of what these fibers look like under stereographic projection $S^3\setminus\{\mathrm{pt}\}\to\mathbb{R}^3$ if you do image searches.
