The highest weight $(0,1,0)$ gives the six-dimensional second <exterior power> of the defining representation. Adding each unordered pair of the four defining weights from part i gives
$$
\begin{gathered}
(0,1,0),\qquad
(1,-1,1),\qquad
(1,0,-1),\\
(-1,0,1),\qquad
(-1,1,-1),\qquad
(0,-1,0).
\end{gathered}
$$
There are no multiplicities, in agreement with the assumption in the question.

Solution

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

 Solution

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