Past exam of the groups and representations course of the University of Oxford / 2018 / 4 / c

For the branching from B3 to A3, apply the supplied weight projection matrix to column vectors of Dynkin labels. It sendsIt is consistent with choosing the $A_3$ simple roots in the order ${\beta }_1=e_2-e_3$, ${\beta }_2=e_1-e_2$, ${\beta }_3=e_2+e_3$. Their Dynkin diagram is the chain ${\beta }_1-{\beta }_2-{\beta }_3$. Their coroot pairings are respectively $a_2,a_1,a_2+a_3$, explaining the projection without confusing labels and Euclidean coordinates.

The seven weights in b project asThe six nonzero weights form the weight system of the SU(4) two-index antisymmetric representation of highest weight $(0,1,0)$. The zero weight is the remaining singlet representation. ConsequentlyThe eight spinor weights project asThe four weights $(1,0,0),(-1,1,0),(0,-1,1),(0,0,-1)$ constitute the defining fundamental representation $4$. The other four are their negatives, forming the dual representation $\overline{4}$ of highest weight $(0,0,1)$. HenceFor the Young tableaux and tensors, let $U={\mathbb{C}}^4$ be the defining SU(4) representation. The singlet representation is ${\Lambda }^{0}U$, with the empty Young diagram and a scalar. The other three Young diagrams are columns of one, two and three boxes, representing $U$, ${\Lambda }^{2}U$ and ${\Lambda }^{3}U$:The square brackets indicate tensor index antisymmetrization, with indices from 1 to 4. The invariant Levi-Civita symbol identifies the three-index antisymmetric tensor with a covector, $w_{\ell }={ϵ}_{\ell ijk}{T}^{ijk}/3!$, explaining ${\Lambda }^{3}U\cong U^{\ast }$.