The diagonal <Cartan subalgebra> consists of
$$
\operatorname{diag}(t_1,t_2,t_3,-t_1,-t_2,-t_3).
$$
Let $\varepsilon_i\in\mathfrak t^*$ read off $t_i$. The roots are
$$
\boxed{
\Phi=\{\pm\varepsilon_i\pm\varepsilon_j:
1\leq i<j\leq3\}},
$$
the <D3 root system>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 102 / 3 / b / i / Solution

{c}
{title2=$D_3$}

The $D_3$ roots are $\pm\varepsilon_i\pm\varepsilon_j$ for $1\leq i<j\leq3$. The simple roots
$$
\alpha_1=\varepsilon_1-\varepsilon_2,\quad
\alpha_2=\varepsilon_2-\varepsilon_3,\quad
\alpha_3=\varepsilon_2+\varepsilon_3
$$
form the three-node chain $\alpha_2-\alpha_1-\alpha_3$, so $D_3\cong A_3$.

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