Lifting elements of $SO(3)$ to $SU(2)$.

Question

Let $A$ an element of ortogonal group $SO(3)$ such that the orders of $A$ is $>2$. We have that $SU(2)$ is a $2$-fold cover of $SO(3)$: $$ \mathbb{Z}_2 \to SU(2) \to SO(3) .$$ So how can I build a lift of $A$ to an element $\tilde{A}$ is $SU(2)$?

Answer 1

Let $\mathbb{H}$ be the quaternions and $q \in \mathbb{H}$ can be represented by $q=a+bi+cj+dk$ which can also be written as $q=z_1+z_2j$ where $z_1=a+bi,z_2=c+di$ are complex numbers. Let $\mathbb{H}_1$ be the group of quaternions with norm $1.$ There is a natural isomorphism

$$\mathbb{H}_1 \longrightarrow SU(2)$$

$$z_1+z_2j \mapsto \begin{pmatrix} z_1 & z_2 \\ -\overline{z}_2 & \overline{z}_1\\ \end{pmatrix}$$

Then, we can identify $\mathbb{R}^3$ with pure quaternions $bi+cj+dk$ and define the action of $\mathbb{H}_1 \cong \text{SU}(2)$ on $\mathbb{R}^3$ by

$$q \cdot q_0 \mapsto q \cdot q_0 \cdot q^{-1}, \;\; q \in \mathbb{H}_1, q_0 \in \mathbb{R}^3$$

Now, because the quaternion norm is multiplicative and coincides with Euclidean norm on $\mathbb{R}^3,$ thus $p: \mathbb{H}_1 \to \text{O}(3)$ is a well-defined group homomorphism.

Moreover, write $q \in \mathbb{H}_1$ as $q=\cos \theta+ \sin \theta q_1,$ where $q_1$ is a pure quaternion of norm $1.$ It is also immediate that the action of $q$ on $\mathbb{R}^3$ is the rotation defined by the axis $q_1$ and the angle $\theta.$ Using the definition of $SO(3)$, we can define a surjective homomorphism $p: \mathbb{H}_1 \to SO(3)$ and as you said, the kernel of $p$ is the center of $SU(2)$ which is $\{\pm 1\}.$