= SU4 adjoint branching to SU3
{c}
{title2=$\mathbf{15}\downarrow_{SU(3)}=\mathbf8\oplus\mathbf3\oplus\overline{\mathbf3}\oplus\mathbf1$}
For the upper-left <SU(3)> <subgroup> of <SU(4)>, the defining <representation> restricts as $\mathbf4=\mathbf3\oplus\mathbf1$. Write a <traceless matrix> in three-plus-one block form. Its upper-left traceless block is the <SU(3)> <Adjoint representation>, its upper-right column is $\mathbf3$, its lower-left row is $\overline{\mathbf3}$, and the remaining diagonal direction $\operatorname{diag}(1,1,1,-3)$ is fixed. These <invariant subspaces> have <dimensions> $8,3,3,1$ and exhaust the <SU(4)> <Adjoint representation>. This also proves the formula by restricting $\mathbf4\otimes\overline{\mathbf4}-\mathbf1$.
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