Past exam of the mathematics course of the University of Cambridge / 2026 / ii / Paper 3 / 19F / a / Solution

$SU(2)=\{A:A^{\ast }A=I,\det A=1\}$; every element is $\left(\begin{array}{cc}a& b\\ -\bar{b}& \bar{a}\end{array}\right)$ with $|a{|}^2+|b{|}^2=1$, giving $S^3$. A maximal torus is $T=\{\mathrm{diag}(e^{it},e^{-it})\}$. A noncentral conjugacy class is determined by trace $2\cos t$, has stabilizer $T$, and is $SU(2)/T\simeq S^2$; it meets $T$ at the two inverse diagonal elements. Since $SU(2)$ has no nontrivial continuous characters, $\det \rho =1$.