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

$$\DeclareMathOperator{\SU}{SU}$$ $$\DeclareMathOperator{\SO}{SO}$$ 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 $$n$$?

• The rest of the sentence says "...corresponding to rotation about $\text{Im}(q)$." About any axis you can consider the one-parameter subgroup of $SO(3)$ or $SU(2)$ which corresponds to rotations about that axis; those are the copies of $S^1$ I mean. Dec 15, 2017 at 1:17
• That makes sense for a homomorphism of $\mathbb{R}$ into $SO(3)$, with kernel $\mathbb{Z}$, you send $x$ to the rotation in the plane orthogonal to $\textrm{Im}(q)$ by $2 \pi x$ radians. I don't understand how the same reasoning can be applied to finding a homomorphism of $\mathbb{R}$ into $\SU(2)$.
– D_S
Dec 16, 2017 at 22:07
• The copy of $S^1$ inside $SU(2)$ being referred to is $\theta \mapsto \exp (h \theta)$. This has period $2 \pi$ because $h^2 = -1$. Dec 18, 2017 at 1:24

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 $$q = \cos(\theta) + \bf{h}\sin(\theta) = e^{\bf{h}\theta}$$ 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 $$\eta(\phi) \equiv \cos(\phi) + \bf{h}\sin(\phi)$$ 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 $2\theta$.