Lifts of rotations in $\operatorname{SO}(3)$ to rotations in $\operatorname{SU}(2)$ Let $R_z(\theta)$ be a rotation matrix around $z$ axis, where $R_z(\theta)\in\operatorname{SO}(3)$. I know that $\operatorname{SU}(2)$ is a double cover of $\operatorname{SO}(3)$, so $R_z(\theta)$ has to pre-images in $\operatorname{SU}(2)$. But how can I found them?
 A: We have $\mathfrak{so}(3)\cong\mathfrak{su}(2)$, see here or Masahito Hayashi, Group Representation for Quantum Theory, 2017 (See here, equation (3.27) on page 82). The Pauli matrices $\sigma_1$, $\sigma_2$ and $\sigma_3$ are a basis for $\mathfrak{su}(2)$ with:
$$[\sigma_i,\sigma_j]=2i\varepsilon_{ijk}\sigma_k
\Leftrightarrow
\left[\frac{\sigma_i}{2i},\frac{\sigma_j}{2i}\right]=\varepsilon_{ijk}\frac{\sigma_k}{2i}.$$
The matrices:
$$L_1=\begin{pmatrix}
0 & 0 & 0 \\
0 & 0 & -1 \\
0 & 1 & 0
\end{pmatrix},\;
L_2=\begin{pmatrix}
0 & 0 & 1 \\
0 & 0 & 0 \\
-1 & 0 & 0
\end{pmatrix},\;
L_3=\begin{pmatrix}
0 & -1 & 0 \\
1 & 0 & 0 \\
0 & 0 & 0
\end{pmatrix}$$
known in quantum mechanics to describe angular momentum for the angular momentum quantum number $l=1$ with $m_l=-1;0;+1$ are a basis of $\mathfrak{so}(3)$ with:
$$[L_i,L_j]=\varepsilon_{ijk}L_k.$$
Therefore the isomorphism is given by the identification $\frac{\sigma_j}{2i}\leftrightarrow L_j$.
Write $\vec\sigma$ and $\vec{L}$ for the collection of the three respective matrices into a single vector, then using a positive scalar $\theta$ and a unit vector $\vec{n}$, we have (See here and here):
$$e^{i\theta\vec{n}\cdot\vec\sigma}
=1\cos(\theta)+i\vec{n}\cdot\sigma\sin(\theta)
\in\operatorname{SU}(2)
\tag{1}$$
$$e^{i\theta\vec{n}\cdot\vec{L}}
=1+\underbrace{2\sin\left(\frac{\theta}{2}\right)\cos\left(\frac{\theta}{2}\right)}_{=\sin(\theta)}(\vec{n}\cdot\vec{L})
+2\sin\left(\frac{\theta}{2}\right)^2(\vec{n}\cdot\vec{L})^2
\in\operatorname{SO}(3)
\tag{2}$$
The matrix in equation (2) is exactly the matrix describing the rotation by the angle $\theta$ around the unit vector $\vec{n}$ (which in your case is $e_z$ and simplifies the formula). Given a matrix in $\operatorname{SO}(3)$, $\vec{n}$ can therefore be obtained as its eigenvector and $\theta$ by putting a vector perpendicular to $\vec{n}$ into the matrix. If we have $\theta$ and $\vec{n}$, we can put them into equation (1) to compute a preimage in $\operatorname{SU}(2)$ unter the covering and simply by just swapping the sign, getting the other.
It should be noted that this results in $\operatorname{PSU}(2)=\operatorname{SU}(2)/\{\pm 1\}$ being homeomorphic to $\operatorname{SO}(3)$ (See here).
