The matrices are rotations about the third coordinate axis. Direct multiplication gives
$$
g(\theta)g(\varphi)=g(\theta+\varphi\bmod 2\pi),
\qquad g(0)=I,
\qquad g(\theta)^{-1}=g(-\theta),
$$
and $g(\theta)^{\mathsf T}g(\theta)=I$. Thus they form a <one-parameter subgroup> of the <orthogonal group> $O(3)$, isomorphic to the <circle group>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 302 / 1 / a / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/iii/paper-302/1/a/solution