# 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?

• Use the map here. Commented Aug 5, 2022 at 11:35
• @DietrichBurde I'm familiar with this map. But I didn't succeed to get a lift to rotations. I thought maybe there is a more simple way or answer to this... Commented Aug 5, 2022 at 11:37

## 1 Answer

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).