Branching of a representation

The branching of a group representation $V$ to a subgroup $H$ is its decomposition into irreducible representations of $H$, when that restriction of a representation is a semisimple representation: $V{|}_H={⨁}_{\lambda }{m}_{\lambda }{W}_{\lambda }$. The nonnegative integers $m_{\lambda }$ are the branching multiplicities. For finite groups over the complex numbers, they are the character inner products of the restricted character with the irreducible characters of $H$. Restriction to an arbitrary subgroup need not be completely reducible.

For the embedding $U\mapsto \mathrm{diag}(1,U)$, the defining SU(6) representation restricts to the defining SU(5) representation plus a trivial representation. Taking dual representations, symmetric squares, exterior squares and traceless endomorphisms gives $\overline{6}\downarrow =\overline{5}\oplus 1$, $21\downarrow =15\oplus 5\oplus 1$, $15\downarrow =10\oplus 5$, and $35\downarrow =24\oplus 5\oplus \overline{5}\oplus 1$. The fixed diagonal direction in the Adjoint representation is $\mathrm{diag}(-5,1,1,1,1,1)$. Dimensions label representations of different groups on the two sides: the fifteen-dimensional SU(6) representation is alternating, while the fifteen-dimensional SU(5) representation is symmetric.

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 $4\otimes \overline{4}-1$.