= Solution
Under the stated <SU(2)> embedding, the first two components of the defining <SU(3)> <representation> form a doublet, while the third is fixed. Hence $\mathbf3\downarrow_{SU(2)}=\mathbf2\oplus\mathbf1$. Its <dual representation> restricts in the same way because the <SU(2)> doublet is self-dual.
For the <Adjoint representation>, restrict $\mathbf8=\mathbf3\otimes\overline{\mathbf3}-\mathbf1$. The <SU(2)> <tensor product> $\mathbf2\otimes\mathbf2=\mathbf3\oplus\mathbf1$ gives
$$
\mathbf8\downarrow_{SU(2)}
=(\mathbf2\oplus\mathbf1)\otimes(\mathbf2\oplus\mathbf1)-\mathbf1
=\mathbf3\oplus\mathbf2\oplus\mathbf2\oplus\mathbf1.
$$
Here subtraction denotes removal of the <trivial representation> generated by the identity <matrix>; it is not a negative-dimensional summand. The first <SU(3)> factor is unchanged, so the requested <branching of a representation> rules are
$$
\boxed{\begin{aligned}
(\mathbf3,\mathbf3)&\downarrow_H=(\mathbf3,\mathbf2)\oplus(\mathbf3,\mathbf1),\\
(\overline{\mathbf3},\mathbf1)&\downarrow_H=(\overline{\mathbf3},\mathbf1),\\
(\mathbf1,\mathbf3)&\downarrow_H=(\mathbf1,\mathbf2)\oplus(\mathbf1,\mathbf1),\\
(\mathbf1,\mathbf8)&\downarrow_H=(\mathbf1,\mathbf3)\oplus2(\mathbf1,\mathbf2)\oplus(\mathbf1,\mathbf1).
\end{aligned}}
$$
The <dimensions> on the two sides are respectively $9=6+3$, $3=3$, $3=2+1$ and $8=3+2+2+1$.
Back to article page
