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

For $SU(4)$, Dynkin labels $(a,b,c)$ correspond to a Young diagram with row lengths $(a+b+c,b+c,c)$. Numbered boxes below give representative Young tableaux; they specify the shapes, not additional component values:Here the dimensions are $4$, $\left(\genfrac{}{}{0ex}{}{4}{2}\right)=6$, $\left(\genfrac{}{}{0ex}{}{4}{3}\right)=4$, and $4^2-1=15$. The third is the dual representation $\overline{4}$: the invariant volume tensor identifies $b_l={ϵ}_{lijk}{B}^{ijk}/3!$. The last is the Adjoint representation, equivalent to the shape $(2,1,1)$ Young symmetrizer construction after converting its three-box column to one lower index and removing the trace.

For $U\in SU(4)$, the tensor transformation laws areThese preserve antisymmetry and tracelessness. The Adjoint representation can be treated as complex traceless matrices; a real adjoint scalar field is conventionally represented by Hermitian traceless matrices with the same conjugation action.

Adding one box to the one-box Young diagram gives row shapes $(2)$ and $(1,1)$. Thus the first tensor product splits into the symmetric square and exterior square:Their tensors are respectively $S^{ij}=S^{ji}$ and $A^{ij}=-A^{ji}$. Adding one box to the three-box column gives shapes $(2,1,1)$ and $(1,1,1,1)$. The full four-box column is the determinant representation, trivial because $\det U=1$. ConsequentlyFor a mixed tensor $T^i{}_j$, the corresponding pieces are $T^i{}_j-\frac{1}{4}{\delta }^i{}_j{T}^k{}_k$ and $\frac{1}{4}{\delta }^i{}_j{T}^k{}_k$.