Past exam of the groups and representations course of the University of Oxford / 2024 / 4 / d

Let index $0$ denote the first coordinate, which the embedded $SU(5)$ fixes, and let $a,b=1,\dots ,5$ denote the remaining coordinates. The fundamental representation and its conjugate representation immediately giveFor a symmetric tensor, $S^{00}$ is a trivial representation, $S^{0a}=S^{a0}$ is a $5$, and $S^{ab}$ is the fifteen-dimensional symmetric square of a group representation of $5$. For an antisymmetric tensor, $A^{00}=0$, $A^{0a}=-A^{a0}$ is a $5$, and $A^{ab}$ is the ten-dimensional exterior power ${\Lambda }^{2}5$. HenceHere the left-hand $15$ is the antisymmetric tensor of $SU(6)$, while the right-hand $15$ in the first equation is the symmetric tensor of $SU(5)$; equal dimensions alone do not identify representations of different groups.

Finally decompose the traceless block matrixThe traceless part of $D$ is the Adjoint representation $24$, the column $c$ is $5$, the row $b$ is $\bar{5}$, and the one remaining diagonal trivial representation is generated by $\mathrm{diag}(-5,1,1,1,1,1)$. ThusAll dimensions match: $6=1+5$, $21=15+5+1$, $15=10+5$ and $35=24+5+5+1$.