Past exam of the groups and representations course of the University of Oxford / 2021 / 4 / b

Under the two-block subgroup $H_2$, the defining vector space splits into its first and last two coordinates. ThereforeFor the exterior square, the exterior square of a direct sum identity gives two within-block determinant representations, each trivial for $SU(2)$, and the cross-block tensor product $(2,2)$. For the Adjoint representation, use $\mathrm{End}({\mathbb{C}}^4)-1$: each diagonal two-by-two block contributes a scalar and a traceless triplet, while each off-diagonal block gives $(2,2)$ because the SU(2) doublet is self-dual. Removing the overall trace leaves one scalar. The dimension counts are $2+2=4$, $1+1+4=6$, and $3+3+4+4+1=15$.

Under $H_3$, the first three coordinates carry the defining SU(3) representation and the fourth is fixed. HenceIndeed ${\wedge }^2(3\oplus 1)={\wedge }^{2}3\oplus 3=\overline{3}\oplus 3$. The upper three-by-three block of a traceless matrix supplies the Adjoint representation $8$ and one scalar, while the upper-right and lower-left blocks give $3$ and $\overline{3}$. This establishes the branching of a representation in each case without a dimension-only guess.