The <Weyl reflection> $s_{\alpha_1}$ swaps $\varepsilon_1$ and $\varepsilon_2$, whereas $s_{\alpha_2}$ changes the sign of $\varepsilon_2$. Hence
$$
\boxed{
\begin{aligned}
s_{\alpha_1}(\alpha_1)&=-\alpha_1,
&s_{\alpha_1}(\alpha_2)&=\alpha_1+\alpha_2,\\
s_{\alpha_2}(\alpha_1)&=\alpha_1+2\alpha_2,
&s_{\alpha_2}(\alpha_2)&=-\alpha_2.
\end{aligned}}
$$

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 102 / 3 / c / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/iii/paper-102/3/c/solution