Understanding the double cover $SU(2) \rightarrow SO(3)$ via quaternions

Question

$\DeclareMathOperator{\SU}{SU}$ $\DeclareMathOperator{\SO}{SO}$ I'm reading this blog post about the double cover $\SU(2) \rightarrow \SO(3)$. Let $H$ denote the division algebra of real quaternions. Let $\SU(2)$ be the group of $2$ by $2$ complex matrices of the form $\begin{pmatrix} \alpha & \beta \\ -\overline{\beta} & \overline{\alpha} \end{pmatrix}$, where $|\alpha|^2 + |\beta|^2 = 1$. There is an embedding of $\SU(2)$ into $H$ by sending

$$\begin{pmatrix} \alpha & \beta \\ -\overline{\beta} & \overline{\alpha} \end{pmatrix} \mapsto a + bi + cj + dk$$

where $\alpha = a+bi, \beta = c+di$. The image consists of all quaternions with norm one.

Let $q \in \SU(2)$, identified with its image in $H$. We can write $H = \mathbb{R} \oplus \mathbb{R}^3$, where $\mathbb{R}^3$ is the span of $i,j,k$. Conjugation by $q$ stabilizes $\mathbb{R}^3$, preserves the dot product coming from the basis $i,j,k$, and also fixes the vector $\textrm{Im}(q) \in \mathbb{R}^3$, where the imaginary part of a quaternion $a+bi+cj + dk$ is defined to be $bi + cj + dk$. Therefore, $\textrm{Ad}(q)$, the restriction of the conjugation by $q$ map to $\mathbb{R}^3$, lies in $\SO(3)$, the group of rotations in $\mathbb{R}^3$.

Now, let $h = \frac{\textrm{Im}(q)}{|\textrm{Im}(q)|}$. Then $h^2 = -1$, so $h$ is like the imaginary constant. We can write $q = \cos \theta + h \sin \theta$ for some real number $\theta$.

What I don't understand is what is written next about looking at two copies of $S^1$:

What is the copy of $S^1$ inside $\SU(2)$? What is the copy of $S^1$ inside $\SO(3)$? Is this related to the great circle in the unit sphere in $\mathbb{R}^3$ orthogonal to the element $h$? How is a smooth homomorphism $S^1 \rightarrow S^1$ given? And why is $\textrm{Ad}(q)$ then a rotation by $n \theta$ for some $n$?

Answer 1

Qiaochu is correct in his responses. I read the blog post as well, it was well written.

When we select our $q$ from $SU(2)$, or rather our norm-1 quaternion, we are then given $h$, our axis of rotation. Following the blog we arrive at \begin{equation} q = \cos(\theta) + \bf{h}\sin(\theta) = e^{\bf{h}\theta} \end{equation} for some real $\theta$. The copy of $S^1$ in $SU(2)$ is the copy of $S^1$ in $SU(2)$ as a "subspace" of $\mathbb{H}$, explicitly we have elements \begin{equation} \eta(\phi) \equiv \cos(\phi) + \bf{h}\sin(\phi)\end{equation} which are all norm-1 quaternions. The exponential formula gives that this is a homomorphism in $\phi$ mapping $S^1$ to the $SO(3)$ subgroup of rotations about $\bf{h}$, which we identify as another copy of $S^1$.

About the rotation. So we know that $\mathop{Ad}q$ is a rotation about $\bf{h}$ of some angle, in fact $\eta(\phi)$ are all rotations of varying angles. From above though we see that if $\phi = 0, \pi$, then $\eta(\phi) = \pm1$ which make trivial adjoint actions, and thus must be $\mathbb1 \in SO(3)$. Since only elements of the real subspace of $\mathbb H$ will have trivial adjoint actions, these two choices of $\phi$ are the only ones, meaning as $\phi$ runs from $0$ to $\pi$, the rotations must run through $2\pi$. So then the mapping $\eta(\phi)$ is $2\to1$ from $S^1 \subset SU(2)$ to the rotations $S0(3)$. Explicitly, a quaternion $q = \cos\theta + \bf{h_q}\sin\theta$ is the rotation about $\bf{h_q}$ of $2\theta$.