Inverse Hopf Map I have been playing around with the Hopf map and projections and have a question about the inverse map. 
The hopf map is defined as $\pi: \mathbb{S^3} \mapsto \mathbb{S^2}$ or$$\pi: r \mapsto ri\bar{r}$$ where $r \in \mathbb{S}^3$ has unit length $i \in \mathbb{H}$. I interpret the product $ri\bar{r}$ as a $180^{\circ}$ rotation about $r \in \mathbb{S}^3$ that moves the point $i$ on the $ijk$ sphere to some other point $p =p_1i+p_2j+p_3k$ on the $ijk$ sphere. If you know the point $p \in\mathbb{S}^2$ then you can derive the rotation axis given by
$$r=\frac{1}{\sqrt(2(1+p_1))}(p_1+1)i+p_2j+p_3k$$
My question is why does multiplying $r$ by the unit circle in the $1i$ direction  describe the fiber of the hopf map given by i.e.
$$\pi^{-1}(p) = r e^{it}$$ where $t\in[0,2\pi]$.
I am not looking for a formal proof. Just some justification or reason why this is so. I know this is a rotation of the $1i$ circle by $r$ but why is this one of the fibers of the Hopf map?
Thanks.  
 A: Every quaternion in $\mathbb{H}$ may be expressed as a combination of a scalar component and a vector component, meaning $q=x+\mathbf{u}$ where $x\in\mathbb{R}$ and $\mathbf{u}\in\mathbb{R}^3$. (One also calls these the real and imaginary parts.) The product of vectors has scalar and vector parts given by the dot product and cross product:
$$ \mathbf{u}\mathbf{v}=\underbrace{-\mathbf{u}\cdot\mathbf{v}}_{\mathrm{scalar}}+\underbrace{\mathbf{u}\times\mathbf{v}}_{\mathrm{vector}}. $$
This implies $\mathbf{uv}=\mathbf{vu}$ iff $\mathbf{u}\|\mathbf{v}$ (parallel) and $\mathbf{uv}=-\mathbf{vu}$ iff $\mathbf{u}\perp\mathbf{v}$ (perpendicular).
To multiply $(x+\mathbf{u})(y+\mathbf{v})$ just use the distributive property and the above rule.
The norm of a quaternion is given by $|x+\mathbf{u}|^2=x^2+\|\mathbf{u}\|^2$. If $r$ is a unit quaternion, then there is a convex angle $\theta$ such that $r$'s scalar part is $\cos\theta$ and its vector part has magnitude $\sin\theta$, so factor
$$ r=\cos\theta+\sin\theta\,\mathbf{u}=\exp(\theta\mathbf{u}) $$
for some unit quaternion $\mathbf{u}$. (Note $\mathbf{u}$ is just $r$'s vector part normalized, and unit vectors are square roots of negative one so they satisfy Euler's formula just like $i$ does in $\mathbb{C}$.)
If we pick any 3D vector $\mathbf{v}$ in $\mathbb{R}^3$, then the conjugate $r\mathbf{v}r^{-1}$ will be $\mathbf{v}$ rotated around the oriented axis $\mathbf{u}$ by an angle of $2\theta$. Note $r^{-1}=\overline{r}$ for unit quaternions. (This is only a $180^{\circ}$ rotation if $\theta=90^{\circ}$, so when $r$ itself is a pure imaginary quaternion, i.e. a vector).
The Hopf fibration $S^3\to S^2$ is given by $r\mapsto r\mathbf{i}r^{-1}$. The fiber of $\mathbf{i}$ consists of all unit quaternions satisfying $r\mathbf{i}r^{-1}=\mathbf{i}$, or equivalently $r\mathbf{i}=\mathbf{i}r$, so in other words $r$ commutes with $\mathbf{i}$. Since real numbers automatically commute with everything, this is equivalent to the vector part of $r$ commuting with $\mathbf{i}$. Write the vector part of $r$ as $\mathbf{u}_{\|}+\mathbf{u}_{\perp}$ (the components which are parallel and perpendicular to $\mathbf{i}$); since $\mathbf{u}_{\|}$ commutes but $\mathbf{u}_{\perp}$ "anticommutes" with $\mathbf{i}$, the only way $r$ commutes with $\mathbf{i}$ is if its vector part is parallel to $\mathbf{i}$. Thus, its vector part is $\pm\mathbf{i}$, so $r$ is of the form $a+b\mathbf{i}$, but it is a unit quaternion so it is in the complex plane's unit circle $S^1$.
More abstractly, $S^3$ is a group and we have a group action $S^3\curvearrowright S^2$ given by conjugation (applying $r$ to $\mathbf{v}$ gives $r\mathbf{v}r^{-1}$). Picking the point $\mathbf{i}\in S^2$, its stabilizer is $\mathrm{Stab}(\mathbf{i})=S^1$, and so all the fibers of the Hopf map may be characterized: they are cosets of $S^1$. This is a special case of the orbit-stabilizer theorem: if $G\curvearrowright X$ is a group action then the fibers of $G\to\mathrm{Orb}(x)$ given by $g\mapsto gx$ are the cosets of $\mathrm{Stab}(x)$, and the induced map $G/\mathrm{Stab}(x)\to\mathrm{Orb}(x)$ is an "isomorphism" of $G$-sets.
Your formula for the fibers happens to pick particular representatives for the cosets, namely the $180^{\circ}$ rotations around the axes through $\mathbf{i}$ and other points $\mathbf{p}$ in $S^3$ (except it won't work for the antipodal point $\mathbf{p}=-\mathbf{i}$, in which case you can pick any $r=\cos\theta\,\mathbf{j}+\sin\theta\,\mathbf{k}$).
