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

{title2=$\Lambda^kV$}

The $k$th exterior power $\Lambda^kV$ is the quotient of $V^{\otimes k}$ that makes every tensor with two equal factors zero. It represents alternating $k$-linear maps and has dimension $\binom{\dim V}{k}$ in finite dimensions.


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