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

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/iii/paper-302/3/e/solution