For equal-length candidate roots, the required inner products are
$$
\alpha_1\mathbin\cdot\alpha_2=-1,
\qquad
\alpha_2\mathbin\cdot\alpha_3=-1,
\qquad
\alpha_1\mathbin\cdot\alpha_3=0,
\qquad
\lVert\alpha_i\rVert^2=2.
$$

Set A is invalid because $\alpha_1\mathbin\cdot\alpha_3=-1$ rather than zero; indeed its three vectors sum to zero and are not linearly independent. Set B has all the displayed inner products and is linearly independent, so it is valid. Set C is the standard realization
$$
e_1-e_2,
\quad e_2-e_3,
\quad e_3-e_4
$$
of the <A3 root system> and is also valid.

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 302 / 3 / e / Solution

{c}
{title2=$A_3$}

The $A_3$ root system is $\{e_i-e_j:1\leq i\ne j\leq4\}$ in the hyperplane of $\mathbb R^4$ whose coordinates sum to zero. It has $12$ roots and simple roots $e_1-e_2,e_2-e_3,e_3-e_4$.


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