Under the subgroup in part (b), $(v,s)\in\mathbb R^4\oplus\mathbb R$ transforms as $(Rv,s)$, hence
$$
\mathbf5\downarrow SO(4)=\mathbf4\oplus\mathbf1.
$$
An element of $\mathfrak{so}_5$ has the unique block form
$$
X=\begin{pmatrix}A&v\\-v^T&0\end{pmatrix},
\qquad A\in\mathfrak{so}_4,quad v\in\mathbb R^4.
$$
Conjugation by $\iota(R)$ sends $(A,v)$ to $(RAR^{-1},Rv)$. Thus
$$
\mathbf{10}\downarrow SO(4)=\mathbf6_{\mathrm{ad}}\oplus\mathbf4,
$$
and the dimensions check as $5=4+1$ and $10=6+4$. These are the <SO5 to SO4 branching> rules.

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 302 / 4 / c / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/iii/paper-302/4/c/solution