SU4 adjoint branching to SU3 ($15{\downarrow }_{SU(3)}=8\oplus 3\oplus \overline{3}\oplus 1$)

For the upper-left SU(3) subgroup of SU(4), the defining representation restricts as $4=3\oplus 1$. 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 $3$, its lower-left row is $\overline{3}$, and the remaining diagonal direction $\mathrm{diag}(1,1,1,-3)$ 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 $4\otimes \overline{4}-1$.