Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 302 / 3

On the branch through the identity,Direct calculation gives $M^{†}M=I$ and $\det M=1$. Thus these matrices lie inproviding the three requested matrix groups.

A vector transforms as $x^{\prime a}=R^a{}_b{x}^b$. The matrix fixes $n$, preserves lengths and orientation, and rotates the perpendicular plane through $\theta$. It is a rotation by angle $\theta$ about the axis $n$, with orientation fixed by the epsilon-term convention.

Sinceconjugation by $A\in SU(2)$ preserves Hermiticity, tracelessness, and the determinant. Hence $A(x\cdot \sigma )A^{†}=x^{\prime }\cdot \sigma$ with $|x^{\prime }|=|x|$. Thus $\boxed{G=SU(2)}$.

From $A{\sigma }_b{A}^{†}=R^a{}_b{\sigma }_a$ and
$\mathrm{tr}({\sigma }_a{\sigma }_b)=2{\delta }_{ab}$,

The matrix isIts image is the rotation through $\theta$ about

The other element is $\boxed{-M}$. Conjugation by $M$ and $-M$ is identical, so the homomorphism $SU(2)\to SO(3)$ has kernel $\{\pm I\}$ and is the Spin group double cover.