How is $j=1/2$ representation, $U(R(\theta,\hat{\bf n}))=e^{i{\sigma}\cdot{\hat {\bf n}}\theta/2}$, is a projective representation of ${\rm SO}(3)$?

Question

A projective unitary representation of ${\rm SO(3)}$ satisfies $$U(R_1)U(R_2)=e^{i\phi(R_1,R_2)}U(R_1R_2)\tag{1}$$ where $R_1,R_2\in {\rm SO(3)}$. How to show that the $j=1/2$ representation, $U(R(\theta,\hat{\bf n}))=e^{i{\sigma}\cdot{\hat {\bf n}}\theta/2}$, is a projective representation of ${\rm SO}(3)$ i.e., satisfies the condition $(1)$. To do this, one has to show that for $$R_1R_2=R_3\Rightarrow U(R_1)U(R_2)=e^{i\phi}U(R_3).\tag{3}$$ Any suggestions how to show this or at least check this?

Answer 1

1. OP describes projective representations in terms of a 2-cocycle, see section 3 below. An alternative description is in terms of a quotient $$PSU(2)~:=~ SU(2)/\mathbb{Z}_2~\cong~SO(3),\tag{A}$$ where $SU(2)$ denotes the 2-dimension $j=1/2$ non-projective defining/fundamental/spinor representation and $$\mathbb{Z}_{2}~\cong~\{\pm {\bf 1}_{2 \times 2}\}.\tag{B}$$ In other words, in this latter description the 2-dimensional representation of $SO(3)$ is double-valued, i.e. there are 2 branches $\pm U$ represents the same $SO(3)$ rotation.

2. Let $\vec{\alpha}=\theta\hat{\bf n}$ be a rotation-vector in the axis-angle representation $(\hat{\bf n},\theta)$. The opposite branch is given by the axis-angle representation $(-\hat{\bf n},2\pi\!-\!\theta)$. To describe a general $SO(3)$-element ($SU(2)$-element) it is enough to consider a rotation-vector $\vec{\alpha}\in \mathbb{R}^3$ with length $|\vec{\alpha}|\leq \pi$ ($|\vec{\alpha}|\leq 2\pi$), respectively. Note that the $4\pi$-periodicity of $SU(2)$ becomes the familiar $2\pi$-periodicity of $SO(3)$. See also e.g. this & this related Phys.SE posts.

3. From the non-projective defining representation of $SU(2)$, we have $$U(\vec{\gamma})~=~U(\vec{\alpha})U(\vec{\beta}) ,\tag{C}$$ cf. e.g. this Phys.SE post. As mentioned before, we may assume that $|\vec{\alpha}|,|\vec{\beta}|,|\vec{\gamma}| \leq 2\pi$. However, if we only want to use rotation-vectors with lengths $\leq \pi$ (corresponding to $SO(3)$-rotations), we might have to use the opposite branch. Such a transition costs a non-trivial 2-cofactor in eq. (C).

References:

1. G 't Hooft, Introduction to Lie Groups in Physics, lecture notes; chapters 3 + 6. The pdf file is available here.

Answer 2

Take $\hat{\bf{n}}_1=\hat{\bf{n}}_2$, and $\theta_1+\theta_2=2\pi$. As an $SO(3)$ element, you should have $U(R_1)U(R_2)=U(2\pi)=1$ but here you get $-1$.

Thus, you get $U(R_1)U(R_2)=e^{i\phi}1$, where $e^{i\phi}=-1$ (or $\phi=\pi$)