Unit Quaternions on the 3-sphere, $S^3$ as orthogonal transformations. I am reading through Andrew Hanson's "Visualizing Quaternions" and came across this passage on page 50:

$q(\theta, {\bf n}) = \left( \cos\frac{\theta}{2}, {\bf n} \sin \frac{\theta}{2} \right)$ produces the standard rotation matrix ${\bf R} \left( \theta, {\bf n} \right)$ for a rotation by $\theta$ in the plane perpendicular to ${\bf n}$, where ${\bf n} \cdot {\bf n} = 1$. $\theta$ is an angle obeying $0 \leq \theta < 4 \pi$ rather than $0 \leq \theta < 2 \pi$. This extension of the range of $\theta$ allows the values of q to reach all points of the hypersphere $S^3$.

Here, $q$ represents the quaternion and ${\bf n}$ the rotation axis.
To describe how I visualize $S^3$, I will first describe analogously how to visualize $S^2$. Imagine cross sections of $S^2$ as we pass from $z = 1$ to $z = -1$. The cross sections are circles of radius $\sin\beta$, where $0 \leq \beta \leq \pi$ is the usual polar angle in the spherical polar parametrization of $S^2$. If $0 \leq \alpha < 2\pi$ is the azimuth, then $\alpha$ gives a parametrization of every cross sectional circle.
Now, to visualize $S^3$, I draw a set of 2-spheres, the intersections of $S^3$ with the various hyperplanes $w$ = constant, where we pass from $w = 1$ at the North Pole through the equator at $w = 0$, to the South Pole at $w = -1$. $w$ represents the real part of the quaternion. In this case, we identify $w = \cos \frac{\theta}{2}$. The North Pole and the South Pole are degenerate 2-spheres of zero radius. $\sin \frac{\theta}{2}$ represents the radius of every cross sectional 2-sphere. When $\theta = \pi$, $w = 0$, and we are at the equator of $S^3$, and the radius is unity. 
Here is my concern: I see no reason for extending the range of $\theta$ to cover all the points on $S^3$. $0 \leq \theta < 2 \pi$, allows us to reach every point on $S^3$, with ${\bf n}$ being a unit vector.
Is this an error on the part of Hanson?
 A: It is unnecessary to make $\theta$ range over the interval $[0,4\pi)$ in order to cover all of $S^3$. On the other hand, when you mentally picture rotations around an axis as an animation, you are essentially visualizing a one-parameter subgroup, and for a fixed axis $\mathbf{n}$ in order to parametrize the entire subgroup from the identity and back to itself (in exactly one trip) while we only need $\theta$ to range in $[0,2\pi)$ if we are in the rotation group $\mathrm{SO}(3)$, we need $\theta$ to range in $[0,4\pi)$ if we are in the spin group $\mathrm{Spin}(3)=S^3$ that double covers it. This is likely the explanation for the mixup.
Take $\mathbb{R}^2$ for a simpler example. If you parametrize the double cover of the unit circle (which is again just the unit circle) via $(\cos(\theta/2),\mathbf{n}\sin(\theta/2))$ where $\mathbf{n}\in\{\pm i\}$ we only need $\theta\in[0,2\pi)$, but in order to parametrize the whole circle group $\mathrm{Spin}(2)$ for a fixed choice $\mathbf{n}=+i$ we need to use the double-size interval $[0,4\pi)$.
