= Solution
Let $v_{1,+}=e_1+ie_2$, $v_{1,-}=e_1-ie_2$, $v_{2,+}=e_3+ie_4$, $v_{2,-}=e_3-ie_4$ and $v_0=e_5$. Direct multiplication by the printed rotation blocks gives their torus eigenvalues
$$
e^{ix_1},\ e^{-ix_1},\ e^{ix_2},\ e^{-ix_2},\ 1.
$$
Thus the differentiated <weights of a representation> are $ix_1,-ix_1,ix_2,-ix_2,0$, with this sign convention fixed by the given matrix blocks.
A wedge of two distinct eigenvectors has the product eigenvalue, so its differentiated weight is the sum. The ten vectors in a <basis> of the <exterior square> split into three groups:
$$
\begin{array}{c|c|c}
\text{vectors}&\text{weights}&\text{number}\\
v_{1,\sigma}\wedge v_{2,\tau}&i(\sigma x_1+\tau x_2),\ \sigma,\tau=\pm1&4\\
v_{j,\sigma}\wedge v_0&i\sigma x_j,\ j=1,2,\ \sigma=\pm1&4\\
v_{1,+}\wedge v_{1,-},\ v_{2,+}\wedge v_{2,-}&0&2
\end{array}
$$
The eight nonzero weights are distinct, so each has multiplicity one; the zero <weight space> has dimension two. The multiplicities add to $\binom52=10$, accounting for the entire representation.
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