Conjugation differentiates to the <Adjoint representation>
$$
\operatorname{Ad}_gX=\left.\frac d{dt}\right|_{0}g\exp(tX)g^{-1}.
$$
Because conjugations compose, $\operatorname{Ad}_{gh}=\operatorname{Ad}_g\operatorname{Ad}_h$ and $\operatorname{Ad}_e=I$, so this is a group representation.

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 302 / 2 / b / i / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/iii/paper-302/2/b/i/solution