For $A\in O(3)$ and $X^{\mathsf T}=-X$,
$$
(AXA^{\mathsf T})^{\mathsf T}=AX^{\mathsf T}A^{\mathsf T}=-AXA^{\mathsf T},
$$
so $\operatorname{Ad}_A$ maps $\mathfrak o(3)$ into itself. Moreover,
$$
\operatorname{Ad}_{AB}(X)
=ABX(AB)^{\mathsf T}
=A(BXB^{\mathsf T})A^{\mathsf T}
=(\operatorname{Ad}_A\circ\operatorname{Ad}_B)(X).
$$
It also sends the identity to the identity linear map, so $A\mapsto\operatorname{Ad}_A$ is the <Adjoint representation of a Lie group>.

Solution

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

 Solution

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